# The Great Calculation According to the Indians, of Maximus Planudes

## Introduction

Maximus Planudes was born around 1255 in Nicomedia and died at Constantinople around 1305. He took the name Maximus, replacing his baptismal name of Manuel, when he became a monk, shortly before 1280. Apart from translating theological and classical works from Latin into Greek - a good knowledge of Latin seems to have been a rarity among the Byzantines -  he is best known for his editions and commentaries on Greek poetry and drama, as well as for his training of upcoming scholars, such as Manuel Moschopoulos, who continued the important work of preserving, and ensuring the survival of a number of important Greek works.

As with Moschopoulos, who wrote a work on Magic Squares, Planudes had an interest in Mathematics, evidenced by his editions of Aratos' PhainomenaTheodosios ' Sphairica, Euclid's Elements, (Ps-)Iamblichos' Theologoumena Arithmeticae and Diophantos ' Arithmetica.

The present work, The Great Calculation According to the Indians, introduces (i) the (eastern) Arabic form of the Indian numerals, as used in Persia, along with (ii) a detailed exposition of algorithms for addition, subtraction, multiplication and division, both in the decimal system of these numerals and also in the sexagesimal system (iii), whose applications lie in astronomy. Finally, he gives algorithms for the extraction of square roots (iv), to various degrees of accuracy.

The introduction of this numeral system to Europe cannot be traced down to any one person or event, but seems to have occured in various places independently and over a period of time. The earliest known European manuscript containing the first nine Hindu-Arabic numerals dates from 976 and was found in a monastery in northern Spain. Later, Abraham ben Meir ibn Ezra, after a residency in Toledo in the early twelfth century brought some form of the numeral system, (probably the western Arabic variety), to England sometime between 1140 and 1167, replacing the Arabic symbols with Hebrew letters, but maintaining the decimal structure.

Other figures such as Gerbert of Aurillac (c. 945-1003) and John of Hallifax  (Sacrobosco)(c. 1195-56) in France also played their part in the disemination of the new system.

Planudes may have acquired his knowledge of the numeral system and algorithms during his time in Venice, where he was stationed as Ambassador during the reign of the Byzantine Emperor Andronicos II. Venice was at that time, of course, a major trading city and a vital point of contact between the East and West.

## Outline of the Work

In the first section of the Great Calculation, Planudes tries to explain the numeral system and its symbols, in terms of Greek arithmetic, which used monadic numbers, (units), decadic numbers, (tens), and so on. He also introduces the symbol for zero (the cipher), which was generally absent from Greek arithmetic. The algorithm for addition is identical (as we would expect) to our modern algorithm, and involves the notion of carry. He also mentions the old check, known as casting out nines, to verify (the reasonableness of) the answer. For subtraction he gives the two standard algorithms: the borrow and pay back algorithm and the algorithm often referred to in modern Primary school books as the trading method. These are explained in great and often unnecessary detail with general descriptions of the methods followed by worked examples. The examples chosen by Planudes are not always the best - on several occasions in the work a poor choice of digits leads to confusing ambiguity in the written explanations.

Planudes' first method of long multiplication is rather different from that used by the modern school child. He uses a kind of chiastic, or cross' multiplication, which involves sums of products, combined with carry. For example, to find 24×35 he works out 4×5 = 20 and carries 2. Then he finds 4×3 + 2×5 = 22 and adds on the 2 to obtain 24. He again carries 2 and works out 2×3 which becomes 8 with the carry, so the answer is 840. For larger numbers, the process is similar. For example, to find 432 ×264, we have to find the products 2×4, 3×4 + 2×6, 4×4 + 3×6 + 2×2, 4×6 + 3×2 and finally 4×2, remembering to record the units and carry the tens as we go.

Planudes then gives a second method which involves writing and then erasing numbers. The explanation of this method is complicated and made worse by the fact that the erasure of symbols makes it very difficult to follow his examples. His explanation of long division is particularly difficult to follow.

The third section introduces the sexagesimal system using signs of the zodiac, (that is, multiples of 30° which are read modulo 12), and degrees, minutes and seconds, which are all base 60. The four operations are repeated using this new system.

Finally, there is a long section on finding square roots of non-squares to various degrees of accuracy, based essentially on the formula Ö{a2+e} » a + [(e)/(2a)]. The latter parts of this section are particularly obscure and hard to make sense of.

Planudes' algorithm for square roots is very similar to the modern one' which goes as follows:

Take, for example, 235. Divide the digits from right to left into pairs, thus 2|35. Find the root of the largest square less than the first number, i.e. 1, and subtract its square from the number, i.e. 2-12=1. Record the root and carry down the next number 35, so we have

 1
 Ö 2|35
 1
_____
 2
 1 35

Also, double the square root 1 and write it on the left beside the number 135. We now find the largest digit x such that 2x (i.e. 20 +x) times x is less than 135. The digit is 5, since 25×5 = 125 < 135. We write the 5 on top and subtract 135- 125 = 10.

 1 5
 Ö 2|35
 1
_____
 25
 1 35
 1 25
_____
 10

Then the square root of 235 is 15[10/30] = 15[1/3], noting that the denominator is twice the root 15.  In longer examples, Planudes makes the description more complicated by not properly using place value.

What follows are a series of excerpts from my (fairly literal) translation of the whole work. These excerpts consist of: the introductory passage on the numerals themselves and the section on addition; the first of the two algorithms for subtraction; the first algorithm for multiplication; the introduction to the sexigesimal system and an example of addition in that system; and finally the beginning of the section on square roots.  This should be sufficient to give the general flavour of the work. The full translation can be found at the website:

www.maths.unsw.edu.au/ ~ peter/planudes.pdf

I used the edition of Gerhardt (1865) for the translation, making corrections and additions as necessary. Such alterations are noted when they occur.

There is also an old and very literal German translation by Wäschke listed in the References.

## The Digits of the System, I

The So-Called Great Calculation According to the Indians'

 of the monk Maximos Planoudes.

 A Translation

Since numbers continue without bound, but knowledge of the boundless is not possible, the more eminent of the astronomers invented certain signs and a method relating to them, so that the representation of those numbers they needed might be more easily and more clearly apprehended at a glance.  There are only nine signs required which are these: 1 2 3 4 5 6 7 8 91.  They also use a certain other sign which they call a cipher, which, according to the Indians, signifies nothing'. These nine signs are themselves of Indian origin and the cipher is written as 0.

When each of these 9 signs2 stands alone by itself and in the first place beginning from the right-hand side, the symbol 1 indicates one, 2 indicates two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight and 9 nine. If, however, it is in the second place, then the symbol 1 indicates ten, 2 twenty, 3 thirty and so on. In the third position 1 indicates a hundred,3 2 two hundred, 3 three hundred and so on. The pattern continues for the remaining places.

## The Digits of the System, II

For greater clarity let me also say the following: The number lying in the first place indicates the number of units there are. The second is the number of tens, the third the number of hundreds, the fourth the number of thousands and so on, as far as many places as the number occupies.

Note also that as the number proceeds through the four places it changes its literal name each time.9 Then in the fifth place it takes again its original name, not just the number itself, but tied to its place value. This continues 'till the eighth place in which it takes the name of the fourth, and so it continues in turn. Thus, in the previous example given above, 2 indicates and is read as two', 9 as ninety', 5 as five hundred' and 4 as four thousand'. The sign 7 is then seven (units of) myriads', just as we referred to two' (units) in the first position. Similarly the next sign is for seven, but is in fact seven myriads10, 2 is for twenty myriads, just as we had ninety in the second position, so here the sign means twenty, for both numbers are decadic. This is the same as the case with the monadic numbers that preceeded them, and so on in turn.

The cipher is never placed at the left-hand end of the digits but can appear in the middle of the number or at the right-hand side, that is, at the extreme side before the smallest (non-zero) place digit.11 Not only one, but two, three, four or as many zeros as are required may be placed in the middle or in the other aforementioned place. Just as the (number of) places increases the size of the number, so too does the number of ciphers. For example, one cipher lying at the end makes the number decadic, 50 is fifty in fact, two ciphers make it hecatontadic, thus 400 is four hundred, and so on in turn. If one cipher lies in the middle and there is only one symbol before it, it makes that number hecatontadic, thus 302 is three hundred and two, but if there are two such signs, the number is chiliadic, thus 6005 is six thousand and five.   If there is a single cipher with two signs after it, this indicates a chiliadic number, thus 6043 is six thousand and forty three, but if there are two then the number is myriadic, thus 60043 is six myriads and forty three, and so on in turn. To put it simply12, the number is to be understood by the order in which the symbols are placed.

In regard to astronomy we have then the need of six types of operations13, of which the first is that referred to as numeration, 14 and then addition, subtraction, multiplication and division, while the sixth is the extraction of the square root15 of each (given) number. There has already been mention made of the representations, so let us turn our attention to the method of addition.

Addition is the combination of two or more numbers into a sum resulting in one number, for example when we add two and three we make five. It is performed as follows.

Write down the symbols in turn, as many and whichever ones you like, and again underneath these, write as many digits as you like, with the same number of places or more or less. Let them be placed so that the units16 are placed under the units, the tens under the tens and so on in turn. Then add each to each, that is, the first to the first, the second to the second and so on in similar fashion. Write the number resulting from the two above the first column, that is, the number formed from adding the first column above the first column, that resulting from the second column above the second column and so on in turn. If then the total of the numbers added is two or three or four and so on as far as nine, it is written above, as has been said. But if the total is ten then write a cipher, which indicates nothing', take a one instead of ten and combine it with those numbers being next in line to be added. If the total is more than ten, write down that part of its excess over ten above the numbers added, as before, and take one instead of ten and combine it with those numbers about to be added.

To make the explanation clear by example, I give the following diagram.

8  0  3  0   2
5  6  8  7   8
2  3  4  3   3

We begin then with 3 and 7; 3 and 7 gives 10, write the cipher above, and carry17 the unit, which indicates ten.  Again, 4 and 8 gives 12, add in18 also the unit which you carried and we get19 13.  We again take the unit which indicates ten and write the 3 above. Now 3 and 6 gives 9, add in the unit you carried and we get 10.  Write the cipher above and keep the unit; now say  2 and 5 makes 7, add in the unit you held and we get  8 and write this above. Thus the sum of five thousand six hundred and eighty seven and two thousand three hundred and forty three is eight thousand and thirty.

There follows a test20 by which we can learn whether we have performed the addition correctly or not. Regard the values of the signs no longer as monadic, decadic and so on, but take them all as monadic and add (the numbers in) each of the rows individually and look at the number resulting from each. Subtract nine from each, look at what is left for each number, and place what is left at the end of the row which produced it. If the remaining numbers of the two rows add to less than nine, then it is not necessary to subtract (further) from them, but if what is left exceeds this, then again subtract nine and look at the remainder. If it equals the remaining number in the top third row, then one knows that the addition was effected correctly, but if they are not equal then the opposite is true.

For the sake of clarity, let us demonstrate on the previous example.  We say that 8 and 3 is 11, take away 9 leaves [2. Now  5 and  6 is 11, take away  9 leaves  2, 2 and 8 is 10, take away 9 leaves]21 1. And  1 and  7 is  8 and so write  8 in turn in the middle row.   Again 2 and 3 is 5,  5 and  4 is 9, take away 9 and the resulting number 3 remains. Write this in turn after the numbers added together. Now 3 and 8 is 11, take away 9 and 2 is left over. You then have the number equal to the number from the previous check and so the addition is correct.

## On Subtraction or Taking Away

Subtraction from a number, means to take one number from another and look at what is left over. We always either take the lesser number from the greater, so that there is some remainder, for example three from five leaves two, or we subtract equal from equal, where the remainder is nothing; for example when we subtract three from three. It is not possible to take a greater from a lesser number, for it is not possible to take away what is not there.

If you now want to perform a subtraction, you proceed as follows:

Write the digits in turn as many and whichever you like. Below it write the same number of digits or less, but not more. If the number below which is being subtracted has fewer digits then no other condition need be stated. But if it has the same number of digits as the top number then make sure the last digit of the top row is greater than the last corresponding digit of the bottom row. As stated previously, when I speak of the last digit', I mean the one on our left hand side. We need this digit to be greater than the bottom one to ensure that the whole number (on top) is greater than the whole number (on the bottom). This being so, even if the rest of the digits in the bottom number are greater than the digits in the top number, taken one at a time, nevertheless the top number is greater and in no way smaller than the bottom one, that is, than the number being subtracted.

Suppose then that the numbers are written with digits corresponding, units with units, tens with tens and so on in turn. If the first digit22 on the bottom row is less than the first digit of the top, then take the lesser from the greater and write the number remaining above the first digit on the first row. If it is equal, then subtract equal from equal and, as stated above, write the remaining zero above. If it is greater, since we cannot take the greater from the lesser, borrow a unit which signifies ten, from the next digit after it, that is from the second digit of the bottom row. (This digit after the first place occupies the second decadic position.) Having added this ten to the smaller number in the top row, subtract the larger from that total and write the remainder again above the smaller digit. If you can take the second digit on the bottom line, after adding the unit to it - for this unit is again thought of as a unit in respect to the number in the corresponding column, but as ten in respect to the preceeding column, since every number is taken to be ten times the number in the preceeding column, viz; ten compared to the unit, a hundred to ten, a thousand to a hundred and so on - if then you can take away this second digit, after adding to it the unit, then take it away and write the remainder, if there is any, above that second digit on the top row, but if not then write 0. Again, if the second digit on the bottom row, with the unit, is greater than the second digit on the top, again borrow a unit, that is a ten, from the third column on the bottom and add the ten to the second digit on the top and take away the second digit on the bottom, with the unit, and again write the remainder above the second digit on the top. Proceeding thus to the end, you will have the desired result. For the number remaining after you take the whole of the lesser number from the whole of the greater is precisely the number written above the top row.

So that what is being said might become clearer to you in an example, let it be as follows:23

I wish to take 3 from 2 but I can't, since 3 is greater than 2. I add a unit to the 4 next to the 3. I regard the unit as a ten24 and I say 10 and 2 (make 12). I subtract 3 from 12 leaving 9. I write this number above the 2. Again I wish to take 4, along with the unit, from 1 , but am unable. I add a unit to the 8 after the 4 and, regarding this as ten, I say ten and 1 is 11.

5 4 6 1 2

1 8 7 6 9

5 4 6 1 2

3 5 8 4 3

1 1 1 1

I take from 11 the 4 with the unit, that is 5, leaving 6 and I write this above the 1. Using this method I reach the last column and since I can subtract 3 with a unit from 5, I subtract from it 4, that is 3 with a unit, leaving a unit. This number I write above the 5.

With this in mind another example is given, in order to show that if the remainder is zero, it is still written. Beginning in the third place in the accompanying example,

3 1 2 5 6

0 7 0 8 8

3 1 2 5 6

2 4 1 6 8

1 1 1 1

after adding the unit to the digit thus making 2, I can subtract this completely from 2 and nothing is left. I write 0 above the two. I continue in the same way to the end of the diagram.

A further example will show how the calculation is carried out when there are fewer digits on the bottom line.

6 4 5 4 3 2

6 3 8 6 7 4

6 4 5 4 3 2

6 7 5 8

1 1 1 1 1

25

Going to the 4th place, since I am unable to take the 6 with the unit added, that is 7, from the 5 I write a unit next to the units under the 4 and I regard this as 10. Having added this to the 5, I proceed as previously explained. After this I take the unit from the 4 leaving 3 and I write this above the 4. If there are two, three or more digits less in the bottom than the top26 a unit is not written below those digits from there to the end along the top row, but the digits are written as they are, each one written above itself in turn after those digits written above and included with the remainders.27

If you wish to have a check, proceed as follows. Suppose in the last example, we say 8 and 4 is 12 and take the unit from it leaving 2. Write this above the 4.  Again, 5 and 7 is 12, add in the unit you carried and we see 13, take away 10 and again carry the unit leaving 3. Write this above the 7. Again 7 and 6 is 13, adding in the unit makes 14. Then take away 10 and carry the unit leaving 4. Write this above the 6. Again 6 and 8 is 14 and adding in the unit makes 15. Take 10 amd again carry the unit leaving 5. Write this above the 8. Then since we have nothing further from the bottom line of our initial two rows, take the next digit in turn of those left, that is 3, and again add in the unit which you held making 4. Write this above the 3. Now since 6 has nothing to combine with28 nor is there a unit being carried to add to it, place this 6 as it is above itself. Now the topmost row turns out to be the same as the top row of the two initial rows, and the subtraction was performed correctly. Indeed this is always the method by which the check is made, that is, to add the lower of the two initial rows to the row of remainders. Check that the row resulting from this is the same as the top row of the initial two. If so, then the subtraction has been correctly done.

## On Multiplication

Multiplication takes place when we take one number a certain number of times, and from this produce a new number. For example, twice three is six. We took three, as many times as there are units in two, or vice versa, which produces six. Multiplication is performed as follows:

Write down the symbols in turn, as many and whichever ones you like, and again underneath these, another row of digits, with the same number of places or more or less. Do this however in an orderly fashion so that unit is under unit, the tens under the tens and so on in turn. Then multiply the first digit on the top row with the first digit of the bottom row. If the resulting number is less than ten29 then write it above the first, but if a multiple of ten30, such as ten or twenty or thirty and so on, write zero31 and carry as many units as there are tens in the product, but if (the product) is made up of both kinds, that is, units and tens, such as fifteen or twenty four and the like, then write the number of units above; for example five or four, but in regard to the tens, carry as many units as there were tens. Now multiply the first digit of the top line with the second digit of the bottom line and again the first digit of the bottom line with the second digit of the top and take the sum of these products. Add on the monads which you carried, whether two or more, and again, if a number less than 10 results, write it down above the second digit, or if a multiple of ten or mixed, proceed as previously shown. Now if each row has only two digits, there remains only to multiply the second digit by the second digit and then add to this any units that might be carried and write the result after the numbers previously written. If there are three digits, multiply the first by the third and again the first by the third chiastically32 and also the second by the second and adding these up write it down as you were instructed.33 Then multiply the second by the third and the second by the third chiastically, record it and then multiply the third by the third and again record it.
To make the explanation clear by example, I give firstly the following diagram having two rows each with two digits.

8  4  0

2  4

3  5

We say four-times five is twenty and write zero34 above the 4 since twenty is a multiple of ten, and carry the two. We then say four-times three is twelve and five-times two is ten and together they make twenty two. To this add the two units we carried and it becomes 24. We write the 4 above the 2 and carry the two units. Again we say twice three is six and add on the two units and it becomes eight. I then write 8 in turn next to the previous digits and this number is the product of twenty four by thirty five.

Now consider another diagram where the lines have three digits.

1  1  4  0  4  8

4  3  2

2  6  4

We then say twice 4 is eight and write this above the 2. Twice 6 is 12 [and four-times 3 is 12]35 and together they make 24. We write 4 above the 3 and carry 2. Again twice 2 is 4, four-times 4 is 16, and thrice 6 is 18 and together they make 38. We add the two units to this to give 40. Write zero above the 4 and carry 4. Then we say thrice 2 is 6, six-times 4 is 24, total 30. Add on 4 gives 34. Write 4 next to the zero and carry 3. Now we say four-times 2 is 8 and add  3 making 11, and write this in turn after the 4.

This is how the multiplication proceeds if there is an equal number of digits in each of the rows, but if one exceeds the other, fill up the row with the smaller number of digits with zeros and repeat the method outlined. To make this clear by example, we illustrate as follows:

7  6  8  4  2

1  4  2  3

0  0  5  4

Thrice 4 is 12. Write 3above the 2 and carry 1. Also thrice 5 is 15 and four-times 2 is 8, together giving 23. We add on the unit, total 24 and write 4 above the 2 and carry 2. Now thrice zero is zero, four-times 4 is 16, twice 5 is 10 , a total of 26. Add on the two giving 28. We write the 8 above the 4 and carry 2. Also thrice zero is zero, four-times 1 is 4, twice nothing is nothing and five-times 4 is 20, giving 24 . Add on 2 to give 26 . Write 6 above the the 1 and carry 2. Now twice nothing is nothing, five-times 1 is 5, four-times 0 is 0 making 5, and add on 2 gives 7. We write this in turn next to the 6. Then four-times 0 is 0, nothing-times 1 is nothing and (so) we do not write anything, also once 0 is 0 and again I do not write anything.

Observe that when you have multiplied the first digit (on the top row) by the last digit (on the bottom row), you should make some sign on the first digit to indicate that it has been multiplied by all the digits in turn and must not be multiplied again.

## On the Circle of the Zodiac

Let us now make some comments about the zodiac, degrees, first parts, and secondary parts.36 Note then that the sun passes through the 12 signs of the Zodiac as it passes through its cycle. Each of these is divided into 30 degrees and each degree is divided into 60 primary parts and each primary part into 60 secondary parts, and each secondary part into 60 tertiary parts and so on forever. Astronomers ignore all the rest and only concern themselves with 4 quantities, that is, the signs of the Zodiac, degrees, primary parts, which they simply call minutes' and seconds' leaving out the mention of parts' from the seconds' and primary' from the minutes'.

Whenever you wish to perform an addition, you do it as follows: Write down zodiac, degrees, minutes and seconds, whatever you find in the ****37 of Astronomy, but if not, then whatever you like. Below write the number of the zodiac, the number of degrees in their columns and likewise the number of minutes and seconds. For example, the following diagram makes clear what we are saying.

Zodiac   Degrees   Minutes   Seconds
0           22            57            35
3           18            44            52
4           23            54            49
2           25            32            25
1           14            45            29
2           22            12              2
So beginning from the last38 entry, 9 and 5 make 14 and 9 is 23and 2 is 25. Write the 5 above and hold the 2 which we will add to the next column. 2 and 2 is 4 and 2 is 6 and 4 is 10 and 5 is 15. See how many times we need to subtract 6 from 15 and we see it is 2 remainder 3. Write the 3 above the second row, above the 5 and the 2 under the 5 in the minutes column. Say 2 and 5 is 7 and 2 is 9 and 4 is 13 and 4 is 17. Write the 7 above the four and carry one. Again we say 1 and 4 is 5, 5 and 3 is 8 and 5 is 13and 4 is 17. Check again how many times we can take away 6. It is 2 remainder 5. Write the 5 and carry the 2. Take this with the degrees and say 2 and 4 is 6, 6 and 5 is 11 and 3 is 14 and 8 is 22. Write the 2 and carry the 2. Say 2 and 1 is 3 and 2 is 5 and 2 is 7 and 1 is 8. See how many times you can take 3 from 8 and it goes 2 remainder 2. Write the 2 and carry 2, which is twice thirty. Again say 2 and 1 is 3 and 2 is 5 and 4 is 9 and 3 is 12. Take away twelve and zero is left.39 The answer is 22 degrees, 57 minutes and 35 seconds and the sign of Aries40.

## On Finding the Square Root of Any Number

Since we have dealt with different problems arising from astronomical calculation, let us now deal with the quadrature of numbers which are not perfect squares, to show clearly that it is possible to find the square root of any given non-square number. This is achieved in the following way:

Take the square root of the nearest41 perfect square and double it. Then take from the number whose square root you seek, the square you found nearest to it, and express the remainder as a fraction of the number obtained by doubling the square root42 of the square.43 For example, if eight were double the side of the square, take the fraction as eighths, if ten then as tenths, and so on.

Thus, suppose we wish to regard 18 as a square and find its root. Take a root of the perfect square nearest it, that is, of 16; its root is 4. Double this and obtain 8, and subtract 16 from 18 leaving 2; express this in terms of eighths and write that the square root of 18 is 4 and two eighths. Now two eighths equals a quarter and so the root is 4 and a quarter. To see that this in the correct answer, multiply 4 and a quarter by itself and so find the answer is 18.

Here are the steps: We say then that four-times 4 is 16 and four-times two eighths, that is, one quarter four times, equals four quarters equals one. Furthermore [four-times]44 two eighths is one, and one and one is two. Combine this with the 16 and it makes 18. But this method is too simplistic and is incomplete, and it has a lack of accuracy, for this answer is not the root multiplied by itself. If we multiply also a quarter times itself, we get not just 18, but 18 and one sixteenth. In what follows a more accurate method will be outlined, which we claim to be our own discovery, with the help of God. Let us turn our attention to this method for a while, showing how it can be used even for large numbers, so that it might become easier for us to understand.

We apply the aforementioned method to numbers from the unit to 99 and from a hundred to 9999, since the roots of all numbers between these two have two digits, and we should proceed by the following method:

Suppose we write the number 235 and seek to find its square root.

 1 3 2 3 5 1 5 10 2 10

Take then the number which when multiplied by itself exactly produces the first digit two or is nearest to it. 2 is not correct since when multiplied by itself it gives 4. The number produced by the multiplication must be either equal to 2 or less than it. Hence the product should be one times itself, and we say one times one is again one. Subtract this from 2 and one is left over. Write this in small print above in the space between the 2 and the 3....  Now double the one which you found from the subtraction of one from 2. I refer here to the one which was the root not the square, for you should double that number which you found from the subtraction45. One and one is two. Write this two below the 3, but not so that it lies on the same line as the unit previously written below the 2, but below that again in the third row. Now find the number which when multiplied by this cancels with 13, but when multiplied by itself cancels with the residual46. I mean that the numbers must be less than or equal to 13 and the residual. This is what is always meant by cancel'.47 Six multiplied by two can be subtracted from 13, but when multiplied by itself produces a number which is greater than the residual. The remainder is 1348 and the square is 36. Hence we pass over the 6 and take 5 and say five-times 2 is 10.49 Write the 5 between the 3 and the 2 on the same row as the unit written below the 2. Taking the 10 from the 13 leaves 3. Write this in small print above in the space between the 3 and the 5. Again we multiply the 5 by itself giving 25 and subtract from 35 leaves 10.50 Write this outside of the row by itself. Now double the 5 and it becomes 10, just as at the beginning we doubled the unit below the 2. Write this in the third row in turn next to the two you wrote previously. Now combine the 20 and the 10 in the third row, since 2 lies in the tens column, this gives 30. Take half from it and we get 15, since 30 is double the root. The root then of 235 is 15 and ten thirtieths. The remainder is expressed in terms of that same thirtieth which is obtained by doubling the (integer part) of the square root. Now ten thirtieths is a third. Note that the (integer part) of the square root, that is 15, lies in the second row and its double is in the third. To check the result, proceed as follows: Multiply 15 by itself and by ten thirtieths and say, fifteen by 15 is 225, then fifteen times a third, that is ten thirtieths, gives 5 then 5 and 5 is 10. Combining this with 225 gives 235.

## References and Notes

Gerhardt, C.J. Das Rechenbuch des Maximus Planudes., Eislebes, 1865.

Ifrah, G. From One to Zero; A Universal History of Numbers, New York,

Penguin Books, 1985.

Wäschke, H. Das Rechenbuch des Maximus Planudes., Halle, 1878.

Wendel, C. Planudea in Byzantinische Zeitschrift, 40, 1940, pp. 406-445.

Wilson, N.G., An Anthology of Byzantine Prose, 1971, 126-129.

1. For ease of reading, I will use modern forms for the symbols.

2. I will use numerals when they are given in the Greek and names when Greek names for numbers are given.

3. Note that this is one word in Greek, as with the other multiples of a hundred. This will be important in understanding P.'s description of the four types later in the work.

4.  lit. monads'.

5. Greek uses the word muriadeV or myriads' for tens of thousands.

6. It appears that the word muriadwn has dropped out of the text here.

7. I have everywhere replaced the first person singular with the first person plural.

8. i.e. multiples of a thousand.

9. This and the following passage arise from a quirk of the Greek naming of numbers, rather than anything implicit in the number system. P. is saying that there are four types (lit. signs): units, tens, thousands and myriads. After this, we repeat the types as units of myriads, tens of myriads, hundreds of myriads and thousands of myriads. Then, we again repeat the types as units of myriads of myriads, tens of myriads of myriads and so on. The difficulty arises because Greek uses one word for say three thousand but two for three hundred thousand (three tens-of-myriads).

10. lit. seven myriadically'.

11. The Greek is verbose and clumsy, lit. it is placed at the extreme in the direction of the smallest numbers'.

12.  Lit. To speak simply'. Wilson reads 'eipein here for Gerhardt's ¢eipwn.

13.  Gk. sumballomenwn. Representing numbers is not strictly speaking an operation', but there is no simple English equivalent that captures the force of the Greek.

14. Lit. the signs or schemata'.

15. lit. taking the side of each number as though it were a square'.

16. lit. monadics'.

17. lit. take hold of'.

18. lit. take on board', a nautical term.

19. lit. behold'.

20. This is the old method of casting out nines.

21. This section in square brackets has been restored to the text by Gerhardt.

22. Gk. Here the word shmei on sign' is used instead of the the usual schma. I have translated both as digit'.

23. We are here subtracting 35843 from 54612 giving 18769. The carried units are written underneath and the larger number is written again above, which will be used as part of the checking process later.

24. Literally  as decadic'

25. The first 1 on the bottom line appears to have dropped out of the text.

26. Read o katwterw stícoV tou 'anwterw, ... instead of o kat wterw sticoV, twi 'anwterwi.

27. This almost incomprehensible sentence seems to be simply saying that when we have noting to subtract from a digit, we just copy the digit down.

31. lit. nothing

32. that is, in cross formation like the Greek letter chi c

33. i.e. as well of course as adding on any units that were carried

34. lit. nothing.

35. This section in square brackets has been restored to the text by Gerhardt.

36.  Greek: mrwn -divisions, leptwn    prwtwn - primary parts = minutes, (leptwn)    deuterwn - secondary parts= seconds.

37. The Greek word here is illegible and not included in the text. Possibly it is some word for data

38. lit. smallest.

39. Recall that 12 signs is 360°.

40. lit. first constellation of the ram, Greek kriiou.

41. i.e. nearest but less than

42. lit. side.

43. Symbolically, Ö{a2+e} » a +[(e)/(2a )].

44. This seems to have dropped out of the Greek, restored by Gerhardt.

45. This is confusing. He means that you double the root, not the remainder.

46. i.e. 35.

47. Greek   'anairesiV.

48. The text has 15 which I do not understand.

49. Planudes has made the algorithm very complicated by not using place value. An easier may to express what he is doing is the say that we seek the largest integer x so that (20+xx < 135. Here x=6 doesn't work, but x= 5 does.

50.  i.e. 135 - 5×25 = 10.