Difference between revisions of "מלאכת המספר"

Introduction

Table of content

• In the first book we will discuss the seven types of numerical [operations].

• In the second book we will discuss the methods, ratios, problems and answers needed in this science.

• In the third book we will discuss some techniques and premises common to arithmetic and geometry.

• In the first section we will discuss the definition of arithmetic, number, and unit, as well as a few premises needed and the number of the types of numerical [operations].

• In the second section we will discuss the six types of numerical [operations] with integers.

• In the third type we will discuss the seven types of numerical [operations] with fractions and the multiplication techniques of fractions in astrology [= sexagesimal fractions]

Book One: Numbers

Section one: Introduction to Arithmetic – Definitions and Principles

• Definition of arithmetic: arithmetic is a craft that teaches to count many units, their differences and properties that are easily applied by memory.

• Definition of unit: unit is a foundation and the first part of the number, every number consists of it, but it is apart from every number.

• Definition of number: number is defined as a sum of units

The Positional Decimal System

• The numerals

Zero

1) Units

2) Tens

3) Hundreds

4) Thousands

5) Tens of Thousands

6) Hundreds of Thousands

7) Thousands of Thousands

8) Tens Thousands of Thousands

9) Hundreds Thousands of Thousands

10) Thousands Thousands of Thousands

• Every number is either units, or product of ten, or composed from both

Section Two: Integers

Chapter One: Addition

In this species we are able to write as much lines as we wish.

• The units should be written corresponding to the units, the tens corresponding to the tens, the hundreds corresponding to the hundreds and so on, every rank corresponding to its similar.

• Then the numerals are summed as units.

אם שיהיה מאחדים או מעשרות או מורכב משניהם

• The sum of the digits in the rank is equal to units

• The sum of the digits in the rank is equal to tens

• The sum of the digits in the rank is equal to units and tens

5 2 4 3    8 9 6 2

1 4 2 0 5

Chapter Two: Subtraction

• The digit of the subtracted is equal to the digit in the corresponding rank of the subtrahend

• The digit of the subtracted is larger than the digit in the corresponding rank of the subtrahend

• The digit of the subtracted is smaller than the digit in the corresponding rank of the subtrahend:
  

the corresponding digit of the result = the digit of the subtracted + the complement of ten of the subtrahend's digit

then one is added to the succeeding digit of the subtrahend

the same as adding one to loaning one from the succeeding digit of the subtracted

4 2 8 2 2 4 3 2

1 8 5 0

Chapter Three: Doubling

• double the digit is less than ten

• double the digit is equal to ten

• double the digit is more than ten

5 3 7 2

1 0 7 4 4

example: 7; 31; 3

• 6 is the perfect number in the rank of units:

1       2       4       8     1 6     3 2     6 4   1 2 8   2 5 6   5 1 2 1 0 2 4 2 0 4 8 4 0 9 6 8 1 9 2

Chapter Four: Halving

• half the digit is an even number

• half the digit is an odd number

• no preceding digit before the odd digit

• one of the digits in the middle of the number is zero

2 6 2 1 4 4

1 3 1 0 7 2

1 0 4 8 8 7 6

5 2 4 4 3 8

Chapter Five: Multiplication

• the multiplier consists of one digit: one lines - one starting with units

• the multiplier consists of two digits: two lines - one starts with units, the second starts with tens

• the multiplier consists of three digits: three lines - one starts with units, the second starts with tens, and the third starts with hundreds

• The product of units by units is equal to ten

• The product of units by units is more than ten

• The product of units by units is less than ten

• the multiplier consists of one digit: one interim product

• the multiplier consists of two digits: two interim products

• the multiplier consists of more than two digits: more than two interim products

1 2 3 4       1 2

2 4 6 8 1 2 3 4

1 4 8 0 8

1 2 3 4       5

6 1 7 0

והמותר מהפועל נכה אותו במותר הפעול והעולה נשליך ממנו עוד התשיעיו' הנשאר נשמור אותו
והמקובץ מההכאה הראשנה נחלקהו ג"כ בתשעיות ונשמור הנשאר
ואם זה הנשאר או המותר יהיה שוה למותר העולה מהכאת הפועל והפעול ההכאה ההיא אמתית ואם לא אינה אמיתית

• units of the multiplier×tens of the multiplied
• tens of the multiplier×units of the multiplied

וגם כן עשרות הפועל באחדי הפעול אלו שתי ההכאות לבד הם העשות עשרות

• units of the multiplier×hundreds of the multiplied
• tens of the multiplier×tens of the multiplied
• hundreds of the multiplier×units of the multiplied

הא' מאחדי הפועל במאות הפעול
והב' עשרות הפועל בעשרות הפעול
וההכאה השלישית היא ממאות הפועל באחדי הפעול
וקבוץ שלשתם צריך שיתחיל במדרגת המאות

• units of the multiplier×thousands of the multiplied
• tens of the multiplier×hundreds of the multiplied
• hundreds of the multiplier×tens of the multiplied
• thousands of the multiplier×units of the multiplied

הראשנה היא הכאת אחדות הפועל באלפי הפעול
הב' בעשרות הפועל במאות הפעול
הג' הכאת מאות הפועל בעשרות הפעול
הד' אלפי הפועל באחדי הפעול
והעולה מאלו הד' הכאות צריך להתחיל כתיבתו במדרגת אות האלפים

Sums

Shortcuts

Chapter Six: The Sixth Type which is Division

• Example: we wish to divide 144 by 8.

Do as this diagram:

0   0 6 0 1 4 4

1 8

8

• the divisor consists of one digit [= units] alone

The example is that if the divisor is 8 and what remains to divide is 3, we take the ratio of 3 to 8; the ratio is 3-eighths.

Or we convert every 1 of the remaining 3 into 8 parts; they are 24 parts. We divide them by the divisor that is 8; each part receives 3-eighths.

• If the divisor consists of two digits [= units and tens]: we write it beneath the dividend, in the highest rank, leaving a free space, as aforesaid.

• If the divisor consists of three digits: [...] corresponding to the dividend, the first is as units, the second as tens and third as hundreds.

0 0 0   1 2 3 0 9 8 7 6

8 2 3

1 2

Suppose we wish to divide 9876 into 12 parts.

• Since the divisor is of two numerals, we take the two last digits of the dividend, which are 9 and 8 - the 8 are as units and the 9 are as tens, which are 98 above the 12.

We examine how many times the leftmost digit of the divisor, which is 1, can be subtracted from the leftmost digit of the dividend, which is 9. We see that it is 9 times in it.

Indeed, since we say that as many times as the left [digit] of divisor is found in the left [digit] of the dividend, so many times the right [digit] of the divisor should be subtracted from the right [digit] of the dividend, and that is not enough, thus we write 8 as the quotient and subtract it from the 9; 1 remains above the 9.

We say: 8 times 2, which are 16, when we subtract them from 18, 2 remain above the 8 and zero above the 1 of the dividend.

• Now we go back and take the 2 and the 7 - the 2 on the left and the 7 on the right as said - and we arrange them as we arranged the two first digits, saying: how many times the left [digit] of the divisor, which is 1, can be subtracted from the left [digit] of the dividend, which is 2. We find that it can be subtracted twice.

We find also that the right [digit] of the divisor, which is 2, can be subtracted twice from the right [digit] of the dividend, which is 7.

Therefore, we write 2 as the quotient.

We multiply the 2 of quotient by the 1, which is the left digit of the divisor; the product is 2. We subtract it from the 2, which is the left [digit] of the dividend; 0 remains.

We multiply the 2, which is the quotient by the 2, which is the right [digit] of the divisor; the product is 4. We subtract it from the 7; 3 remain above the 7.

• Let 6 be the right [digit] of the dividend and 3 be the left [digit] of the dividend.

We examine how many times the 1, which is the left [digit] of the divisor, can be subtracted from the 3, which is the left [digit] of the dividend. We find that it is 3 times.

We check also if the 2, which is the right [digit] of the divisor, is found so many times in the right [digit] of the dividend. We find that it is.

Therefore, we write 3 as the quotient.

We multiply 3 by 1; the product is 3. We subtract it from the 3 of the dividend; 0 remains.

We multiply also the 3 of the quotient by the right [digit] of the divisor, which is 2; the product is 6. We subtract it from the 6, which is the right [digit] of the dividend; 0 remains. We write 0 above the 6.

00		000
123		1230
9876		9876
823		823
12		12

Section Three of Book One: [Fractions]

Chapter One: Guiding Methods of the Writing Fractions, their Definition, and their Arrangement

As a half, or a third, or a quarter, etc.

Example: when we say 2-thirds, or 3-quarters.

The 2 and 3 are discontinuous quantity, because they state the number and the number is a discontinuous quantity.

The thirds and quarters are continuous quantity, because they state the part, or parts of whichever whole, so they indicate a continuous quantity.

Example: if we wish to write 2-thirds.

We write 2; beneath the 2 a line; and beneath the line 3. As is shown in this diagram:

If you wish to write a half, write it as follows:

The thirds - write like this:

The quarters - like this:

Three-quarters - we write as follows:

Four-fifths- we write as follows:

Chapter Two: Addition of Fractions

• As the one who wants to add or to know the sum of a half, a third, and a quarter.

The number in which all the fractions written in the above diagram are found is 24. It is generated this way: we multiply 2 by 3; the result is 6, and all this, i.e. 6, by 4; it is 24.

In this numbers there are a half, a third, a quarter and it is as their whole.

Because, as one takes from the whole whichever part he wants, so we take whichever part we want from this number 24 that is as a whole.

So, we take the half, the third, and the quarter this way:

We start taking the half this way: we multiply the digit that is above the line of the half, which is 1, by the digit of 3 that is beneath the 1 indicating the third; it is 3. We multiply the 3 by 4 that is beneath the digit indicating the quarter; it is 12, which is a half of the number that we said.

Then, we take the third this way: we multiply the digit of the third that is above the line by the digit of the half that is beneath the line, which is 2; the product is 2. We multiply the 2 by 4; the result is 8, which is a third of this number.

Then, we multiply the 1 that is above the 4 by the 2 that is beneath the line indicating the half; it is 2. We multiply this 2 by 3 that is beneath the line indicating the third; the result is 6, which is a quarter of this number, i.e. the 24.

So, we have three numbers that are a half, a third, and a quarter, which are 12, 8, and 6; their total sum is 26.

The whole, to which the 26 is related, is 24. Therefore, we divide 26 by 24; the result is one integer and 2 remains that cannot be divided. We relate it to 24 and we find that it is one part of 12 of the whole. Hence, sum of the three aforementioned fractions is 1 integer and one part of 12 parts of the whole.

• Suppose we wish to sum up 2-thirds, 3-quarters, 4-fifths, and 5-sixths.

Their diagram is as follows:

The first thing that is needed is that we find the number that is the whole, i.e. in which all these fractions are found. We do that by multiplying all the digits that are beneath the lines like this:

First, we multiply 3 by 4; it is 12. We multiply the 12 by 5; the result is 60. We multiply 60 by 6; the result is 360. This is the number, in which all these fractions are found, and it is the whole that is also called the common denominator.

We should extract all the fractions from it this way:

First, we extract the first fraction in the said diagram, which is two-thirds, this way: we take the 2 that is above the 3 and multiply it by 4; it is 8. We multiply the 8 by 5 that is beneath 4; it is 40. We multiply 40 by 6 that is beneath the 5; the result is 240 and this number is called two-thirds of the common denominator, i.e. the same common denominator that we said is the whole.

Then, we extract the 3-quarters, this way: we take the 3 that is above the 4 and multiply it by all the bottom digits except for the digit that is beneath it, which is 4. As follow: 3 by 3, it is 9; 9 by 5, it is 45; 45 by 6, the result is 270 and this number indicates 3-quarters.

Then, we extract the 4-fifths, this way: we take the 4 that is above the 5 and multiply it by all the bottom digits except for the 5 that is beneath the 4. As follow: we multiply 4 by 3, it is 12; 12 by 4, it is 48; 48 by 6, the result is 288 and this number indicates 4-fifths.

Then, we extract the five-sixths by taking the 5 that is above the 6 and multiply it by all the bottom [digits] except for the 6 that is beneath it, as follows: we multiply 5 by 3, it is 15; 15 by 4, it is 60; 60 by 5, the result is 300 and this number indicates 5-sixths.

We extracted all the said fractions. Now we should sum up all these said fractions and we do this as follows: we take the numbers 240, 270, 288, and 300 and we sum up all; their sum is 1098 and this is the sum of all the said fractions.

We divide all this sum by 360 that is the common denominator, which is the whole as we said; the result is 3 integers and one part of 20 parts of a whole and this is the sum of all the said fractions. Here is its diagram:

• Example: you have already seen in the above diagram that 18 remains that cannot be divided as it is less than the divisor, which is 360. Therefore, we look for the ratio between 18 and 360 and we find that it is the ratio of 1 to 20, so we said above that is is 1 of 20 parts of the whole.

The way to find this ratio is easy: we take these two numbers - 18 above and 360 beneath, and we examine into how many parts the 18 can be divided. We divide the 360 into this number of parts.

• Example: as the 18 is divided into 3 parts and the quotient is 6, so we divide the 360 also into 3 parts and the result is 120.

We divide the 6 again into 3 parts; the result is 2. We divide the 120 into 3 also; the result is 40.

We divide the 2 again into half; the result is 1. We divide the 40 into half also; the result is 20. So, it is one [part] of 20 parts of the whole.

Chapter Three: Subtraction of Fractions

• As seen in this diagram: suppose we want to subtract from three-quarters of the whole two-thirds of the whole and we wish to know how much remains.

We arrange them as follows:

The example of this diagram: the 4 from one side by the 3 from the other side; the result is 12 and this is the number that we call the common denominator. It is called "common denominator", because the two numbers that are beneath the lines take part in it as said.

Now, we should extract 3-quarters of the common denominator this way: we multiply the 3 that is above the 4 by the 3 that is beneath the 2; the result is 9 and these are the 3-quarters of the common denominator, which is 12.

Then, we extract the 2-thirds this way: we multiply the 2 that is above the 3 by the 4 that is beneath the 3; the result is 8 and these are the two-thirds of the common denominator.

First we have found the common denominator, which is the whole.

Second, we have found its 3-quarters, which is 9.

Third, we have found its two-thirds, which is 8.

Since our intention is to subtract two-thirds from 3-quarters, as we have them, we subtract the 8, which is 2-thirds, from the 9, which is 3-quarters; one remains.

We should know what is the one that remains. In order to know this, we divide it by the divisor, or the common denominator, which is 12, then we know what it is. The division method is:

So, it is 1 [part] of 12 parts of the whole.

As seen in this diagram:

The first thing that should be extracted is the common denominator, which is the product of 3 by 12; it is 36.

From this we extract 2-thirds this way: we multiply the 2 that is above the 3 by the 12 that is beneath the 1; it is 24 and these are 2-thirds of the common denominator.

We extract also 1 [part] of 12 parts of the whole this way: we multiply the 1 that is above the 12 by the 3 that is beneath the 2; it is 3 and this is 1 [part] of 12 parts of the whole.

Now, we sum up the 3 with the 24; it is 27.

We divide the 27 by the common denominator, exactly as the way in the first type; it is 27 parts of 36 of the whole.

These parts should be 3-quarters of the whole. To know this, you should examine this diagram:

We do as follows: we multiply the numbers each by its opposite, i.e. 4 by 27 and 3 by 36. If the products are equal it is a proof that this number is 3-quarters of the whole.

Chapter Four: Doubling Fractions

For example: we wish to double 2-thirds, or 5-sixths, or 4-fifths, or any of the fractions we want.

• As the one who wants to double this number, which is 3-quarters.

We double the 3 that is above the line; the result is six. We divide it by 4 that is beneath the line; the result of division is 1 integer and a half, which is double the 3-quarters.

Chapter Five: Halving Fractions

• As the one who wants to halve one-quarter, whose form is this:

We double the 4 that is beneath; it is:

Chapter Six: Multiplication of Fractions

1) Integer or integers alone by fraction alone.

2) Integer or integers and fraction together by fraction.

3) Integer or integers and fraction together by integer or integers alone.

4) Integer or integers and fraction together by integer and fraction together.

5) Fraction by fraction alone

• Example: we wish to multiply 4 integers by 2-thirds.

We do as follows: we write two numbers on both sides this category.

denominator: First, we extract the denominator like this: we take the number that is beneath the line and it is the denominator. It is 3 in this diagram.

numerator: Then, we multiply the 4 integers by 2 that is above the line; it is 8.

We divide this 8 by the denominator that we said it is 3; the result is 2 integers and 2-thirds and this is the product of the said 4 integers by two-thirds.

• As the one who wants to multiply 4 and a half by two-thirds.

denominator: First, we extract the denominator as follows: we multiply the 2 that is beneath the 1 by the 3 that is beneath the 2; the result is 6 and this 6 is the denominator.

numerator: Then, we multiply the 4 integers by 2 that is next to them, beneath the 1; the result is 8. We add to it the 1 that is above the 2; it is 9 and it is the numerator of this side. We multiply the 9 by the 2 of the other side; the result is 18.

We divide this 18 by the denominator, which is 6; the quotient is 3 and this 3 is the product of 4 integers and a half by 2-thirds.

• As the one who wants to multiply 4 and a third by 6.

denominator: Since there is only one type of fractions here, the 3 that is beneath the 1 is the denominator.

numerator: After we write the method of the denominator, we multiply the 4 integers mentioned by 3 that is beneath the 1; it is 12. We add to it the 1 that is above the 3; it is 13. We multiply the 13 by the 6 integers mentioned; the result is 78.

We divide the 78 by 3, which is the said denominator; the result of division is 26 and this is the product of 4 integers and one third by 6 integers.

• As the one who wants to multiply this number:

Example: [we take] the 4 that is beneath the 3 and multiply it by the 2 that is beneath the 1; it is 8 and this is the denominator. We keep it by itself.

numerator: Then, we multiply the 4 integers by the 4 that is beneath the 3 next to them; the result is 16. We add to it the 3 that is above the 4 next to it; the result is 19. We keep it.

We turn to the other side and multiply the 2 integers by the 2 that is beneath the 1 next to them; the result is 4. We add to it the 1 that is above the 2 next to it; the result is 5. We keep it.

So, we have three things: the first is the denominator; the second is the integers of one side that we have converted to the type of the fractions that are next to them and they are 19; the third is the integers of the other side that we have converted to the type of the fractions that are next to them and they are 5.

Now, we multiply the 5 by the 19; the result is 95.

We divide it by the said denominator, which is 8; the result of division is 11 integers and 7-eighths and this is the product that we wanted.

• As the one who wants to multiply 2-thirds by 4-fifths.

denominator: We extract the denominator, which is the product of the digits that are beneath the lines that are 3 and 5; the result is 15. We keep it.

numerator: Then, we multiply the digits that are above the lines that are 2 and 4; the result is 8.

We divide the 8 by the denominator, which is 15. Since the divisor is greater than the dividend, we relate the dividend to the divisor; we find 8 parts of 15 of one integer and this is the product of 2-thirds and 4 fifths.

Chapter Seven: Multiplication of Fractions in Astrology [= Sexagesimal Fractions]

Example: if we multiply seconds by thirds, we say: the decimal value of the seconds is 2 and the decimal value of the thirds is 3; their sum is 5, so the result of their multiplication are fifths.

• Here you have an example of some of them: suppose we want to multiply 2 degrees, 24 minutes, and 43 seconds by 3 degrees, 3 minutes. and 8 seconds.

As seen in this diagram:

Example: if the result of multiplication are 124 seconds, we write 4 seconds in the column of the seconds, and we write the 120 seconds that are 2 times sixty that are 2 minutes in the column of the minutes.

Chapter Eight: Division of Fraction

1) Fraction by fraction.

2) Fraction by integer.

3) Fraction by integer and fraction.

4) Integer and fraction by integer and fraction.

5) Integer by fraction.

6) Integer and fraction by fraction.

7) Integer and fraction by integer.

8) Integer by integer and fraction.

• As the one who wants to divide 3-quarters by 2-thirds.

We write each one next to the other as this:

denominator: First of all, we extract the denominator, by that we multiply 2, which is above the 3, by 4, which is beneath the 3; the result is 8 and this is the denominator.

numerator: Then, we should extract the [numerator], by that we multiply the 3 that is above the 4 by the 3 that is beneath the 2; the result is 9 and this is the numerator.

Next, we divide the numerator by the denominator; the result is 1 and an eighth of a whole and this is the result of division of 3-quarters by 2-thirds that we wanted.

• Suppose we wish to divide 3-eighths by 2 integers.

As this diagram:

denominator: First, we extract the denominator as follows: we multiply the integer, which is 2, by 8, which is beneath the 3; the result is 16 and this is the denominator.

numerator: Then, we extract the numerator, by taking the 3 that is above the 8 and this is the numerator.

We divide this numerator, which is 3, by 16, which is the denominator; the result are 3 parts of 16 parts of one whole and this is what we wanted.

• As the one who wants to divide 2-thirds by 4 integers and a half.

We write each one as in this diagram:

denominator: We extract the denominator as follows: we multiply 4 by 2, which is beneath the 1; the result is 8. We add to it the 1 that is above the 2; it is 9. We multiply the 9 by 3 that is beneath the 2, next to the 1; the result is 27 and this is the denominator.

numerator: We extract the numerator, by that we multiply the 2 that is above the 3 by the 2 that is beneath the 1 of the other side; the result is 4 and this is the numerator.

We divide it by the denominator, which is 27; the result of division is 4 parts of 27 parts of the whole and this is what we wanted.

We write each one as said above.

• As the one who wants to divide 9 and a half by 3 and 4-fifths. Like this:

denominator: We extract the denominator as follows: we multiply 3 by 5, which is beneath the 4; the result is 15. We add to it the 4 that is above the 5; it is 19. We multiply it by 2 that is beneath the 1 in the other side; the result is 38 and this is the denominator. We keep it.

numerator: We extract the numerator by that we multiply 9 by 2 that is beneath the 1 next to it; the result is 18. We add also the 1 that is above the 2 next to it; the result is 19. We multiply this by 5 that is beneath the 4 of the other side; the result is 95 and this is the numerator.

Then, we divide the numerator by the said denominator; the result of division is 2 integers and a half and this is what we wanted.

We write them as said.

• As the one who wants to divide 12 by 4-ninths, like this:

denominator: In this category the denominator is always the digit that is above the line, so the denominator now is the 4 that is above the 9.

numerator: The numerator is the product of the 12 integers by the 9 that is beneath the 4; so the numerator is 108.

We divide it by 4, which is the denominator; it is 27 in the division we wanted.

We write each one as said above.

• Example: we wish to divide 4 and one-third by 5-sixths, like this:

denominator: We extract the denominator, by that we multiply 5, which is above the 6, by 3, which is beneath the 1 next to it; the result is 15 and this is the denominator.

numerator: Then, we extract the numerator by that we multiply 4 by 3 that is beneath the 1 next to it; the result is 12. We add the 1 that is above the 3 next to it; it is 13. We multiply this by 6 that is beneath the 5 of the other side; the result is 78 and this is the numerator.

We divide it by the denominator, which is 15; the result division is 5 integers and one-fifth and this is what we wanted.

We write them as said above.

• We wish to divide 3 and one-third by 4, like this:

denominator: First, we extract the denominator, by that we multiply 4 by 3, which is beneath the 1 in the other side; the result is 12 and this is the denominator in this category.

numerator: We extract the numerator by that we multiply the 3 integers by the 3 that is beneath the 1 next to it; it is 9. We add to it the 1 that is above the 3 next to it; the result is ten and this is the numerator.

We divide it by the denominator, which is 12; the result is 5-sixths and this is what we wanted.

We write each one as said above.

• Example: we wish to divide 50 by two and a half, like this:

denominator: First, we extract the denominator as follows: we multiply 2 by 2, which is beneath the 1 next to it; the result is 4. We add to it the 1 that is above the 2 next to it; the result is 5 and this is the denominator. We keep it.

numerator: Then, we extract the numerator by that we multiply the 50 by the 2 that is beneath the 1 of the other side; the result is 100 and this is the numerator.

We divide it by the denominator, which is 5; the result division is 20 for each one of the two integers of the denominator.

Example: in this category, when we divide 50 by 2 and a half; the result of division is 20 to the whole.

So, we say: if each of the integers receives 20, the half should receive ten, and according to this we divide the fifty.

• Example: you already know that the quotient of the last [example] of this type was 20 and the divisor was two and a half. So, if you multiply two and a half by 20, the result is fifty.

Section Four of the First Book: [Roots]

Chapter One: Giving Guiding Ways for Finding Roots of Square Numbers or Approximate Roots of Non-Square Numbers

The first is the root of a square.

The second is called the root of a cube.

So, 2 is the root of 4, because when it is multiplied by itself it generates four.

Such as 2 times 2; or 5 times 5; or 10 times 10; and so on for the others.

• Example: 4 that has a root is multiplied by 9 that has a root also yields 36, whose root is 6.

• Example: 18 is divided by 8; the result is [2 and a quarter]; its root is one and a half.

If we multiply 18 by 8, the result is 144 and the root of this number is 12, even though 18 and 8 do not have roots.

With the exception of the units, which is the first rank, that has a root that is not generated from other [ranks], but from its product by itself, because one time 1 is 1.

The second rank, which is the tens, does not have a root, because there is no number, whose product by itself generates it.

The third rank, which is the hundreds, has a root, because the product of ten by itself generates 100.

The fourth rank, which is the [thousands], does not have a root, because there is no number, whose product by itself generates it.

The fifth rank, which is the tens of thousands, has a root, because the hundred multiplied by itself generates it, for 100 times 100 yields ten thousand.

The minutes, since they are in the first rank of fractions, do not have a root.

The seconds that are in the second rank of fractions have a root, because they are generated from the multiplication of the minutes by themselves.

The thirds that are in the third rank do not have a root, because they are not generated from the multiplication of others by themselves.

The fourths that are in the fourth rank have a root, because they are generated from the multiplication of the seconds by themselves.

As halves, thirds, fifths, sixths and their like.

As the product of a half by itself that generates a quarter, which is called a half of a half.

The product of a third by itself that generates a ninth, which is called a third of a third.

As the product of a fifth by itself that generates one [part] of 25, which is called a fifth of a fifth.

The tenth, the eleventh, the twelfth, the thirteenth, the fourteenth, and their like do not have a root, for they are fractions of first degree and they are not formed from others.

The tenth of a tenth, the eleventh of an eleventh, the twelfth of a twelfth, the thirteenth of a thirteenth, the fourteenth of a fourteenth, the fifteenth of a fifteenth, and their like have a root.

Such as a half of a half of a half, or a thirteenth of a thirteenth of a thirteenth, and all their like.

Written calculations

If it is odd, write beneath the last rank a number whose product by itself is a number that is equal to the number that is above it, or as close as possible, if an equal number cannot be found.

If [the number of] the ranks is even, one writes beneath the rank that precedes the last rank a number, such that when multiplied by itself yields a number that is equal to the number that is above it, or as close as possible, if a number equal to it cannot be found.

If nothing remains, we write zero above them.

If there is a remainder, what remains is written above the preceding digit, which is the right of the two [digits].

If there are no ranks left, then the first and the third [bottom] digits are the root of the upper number.

• Example: suppose we want to look for the root of a number, which is 5625.

0       2     0 7 0 0 5 6 2 5

7 4 5

Since the [number of] ranks is even, we should begin from the digit that precedes the last, which 6.

• We should write a number, such that when it is multiplied by itself, it yields the same number, or the smallest closest number, when no number is found that is equal to the one that is above, which is 56.

The number that is closest to 56 is 7, because when it is multiplied by itself, it yields 49.

When 49 is subtracted from the number that is above, which is 56, 7 remains.

We write the remaining 7 above the 6, which is one of the digits of 56; and above the 5 we write zero.

We double the 7 that we wrote below as a root; it is 14. We write the ten beneath the 7, erase the 7, and write the 4 in the following rank, beneath the 2.

• Then, beneath the next digit, we write a number, which is 5, such that when it is multiplied by the doubled and by itself, it erases, or consumes the one that is above, as follows:

When multiplying 5 by the one, which is ten of the doubled, the result is 5. When it is subtracted from what is above, which is 7, 2 remains above the 7.

Then, we multiply the mentioned 5 by 4; it is 20. When it is subtracted from what is above, which is 22, 2 remains.

We write 0 above the upper 2 that is tens for the 2 that follows.

Next, we multiply the 5 by itself; we get 25. When it is subtracted from what is above, which is 25, nothing remains.

We receive that the root of the number is 75, as seen in this diagram.

Since nothing remains, we write 0'0' above the 2 and the 5 and this is what we wanted.

Chapter Two: Finding the Roots of Fractions Alone or Fractions and Integers Together

• Example: if the fractions, whose root you are looking for, are quarters, their root is halves.

If they are ninths, their root is thirds.

• As the one who multiplies thirds by themselves that yield ninths, which is a number that is smaller than the thirds that is its root.

• So, in astrology, 3 minutes are not the root of 9 minutes, since the product of minutes by themselves is seconds that are of a different type.

• Example: if they are thirds that do not have a root, we convert them to ninths that have a root.

• Likewise in astrology, if they are minutes and we wish to extract their root, we convert them to seconds and then extract their root.

Chapter Three: Giving one inclusive Method for Finding the Roots of Numbers by Adding Zeros

• Example: if you wrote six zeros and you find that there are more than three digits in the root, which is half the six that is [the number of] the zeros, take the [exceeding ranks] as integers as said and multiply the remaining that are three ranks by sixty.

So, if the preceding product was minutes, the next [product] is seconds and the next product is thirds and so on, until only three zeros remain.

• Example: we wish to know the approximate root of two

We write 2 and add six zeros before it.

1 3 6         2 1 7 0   1 2 4 2 9 2 4 2 0 0 0 0 0 0

1 2 4 8 1 2 4     2 2 8

• First of all, we extract the root in the way that we said in the chapter on roots: we find that its root according to this way is 1414 and a bit more.

• Since there are four digits here, we take the fourth digit as integer, which is one, and we multiply the rest of the digits that are as the number of half the zeros by sixty. The result of multiplication is 24840.

• We take the two last digits, which are 2 and 4, that are more than [the number of] half the zeros and they are minutes. We multiply the remaining 840 by sixty; the result of multiplication is 50400.

• We take what is more than [the number of] half the zeros; it is 50 and this 50 are seconds. We multiply the remaining 400 by sixty; the result of multiplication is 24000.

• We take the 24 that is more than [the number of] half the zeros and they are thirds. Three zeros remain that are half the original zeros that we added.

So, in this procedure we have found that the approximate root of the required number, which is 2, is 1 integer, 24 minutes, 50 seconds, and 24 thirds.

Chapter Four: Giving Guiding Ways for Finding Roots of Cube Numbers or Approximate Roots of Non-Cube Numbers

• As 2, which is a cube root of 8. For, the product of 2 by its square, which is 4, generates 8, which is the cube number of 2.

• Likewise 3 is a cube root of 27, because the product of 3 by its square, which is 9, generates 27, which is the cube number of 3.

• Also 1000 is a cube root of 10, as it is generated from the product of 10 by its square, which is 100.

• For, the first numerical rank, which is the rank of units, has a cube root. Because, the one that is cubically multiplied generates one, which is a cube number.

• The second rank, which is the rank of tens, do not have a cube root, because no number can be found that when multiplied cubically generates ten.

• The third rank, which is the rank of hundreds, do not have a cube root, because no number can be found that generates it when multiplied cubically.

• The fourth rank, which is the rank of thousands, have a cube roots, because it is generated by cubing the ten.

• The fifth rank, which is the tens of thousands, do not have a cube root, for the reason mentioned.

• The sixth rank, which is the hundreds of thousands, do not have a cube root also, for the reason mentioned.

• The seventh rank, which is the thousands of thousands, have a cube roots, because it is generated by cubing the hundred.

Written calculations

Example: if it is 2, we turn it to 6.

• Example: we wish to know the cube root of 12162.

This is its diagram:

0 0 0   0 4 5 2 0 1 2 1 6 7

2 2 6 3

According to what we have said above, there are only two ranks that have cube roots in this number: the units and the thousands, so we start from the rank of the thousands that is the last cube rank of this number, which is 2, and the 1 that is with it.

We write 2 beneath it, since we cannot find a number, [whose cube] is closer to 12 other than 2.

• So, we multiply 2 by itself cubically; it yields 8.

We subtract it from 12; 4 remains above 2 and we write zero above 1.

• Then, we triple the 2; it is 6.

We write it beneath the digit that is third from the 2 that is beneath the upper 6.

We shift the 2 one rank backwards and write it next to the triple, so instead of being third it becomes second to the left.

After the triple, we write 3, whose product by the triple, which is 6, yields 18, and we multiply the 18 by the sub-triple, which is 2; it is 36.

We subtract it from the 41 that is above the 2; 5 remains above the 1 and zero above the 4.

• Then, we multiply the mentioned 3, which is the present digit, by the triple, which is 6; it yields 18. We multiply it by 3; it yields 54.

We subtract it from the digits [that are above], which are 56; 2 remains above the 6 and zero above the 5.

• Then, we multiply the mentioned 3 by itself cubically; it yields 27.

We subtract it from what is above it, which is 27; nothing remains.

So, we write zero above the 2 and above the 7.

Therefore, we know that this number is a [perfect] cube and its root is 23.

• Another example that occurs in a different way: assuming that we wish to know the cube root of 571787.

This is its diagram:

0 0         1 1 0 0   0 5 9 3 2 0 5 7 1 7 8 7

8 8 4 3     2

Since only two ranks that have cube roots are found in this diagram, we start from the last, which is 1.

We write beneath it a number, such that when it is multiplied by itself cubically, it yields a number that is as close to it as possible, which is 8, as there is no closer number there.

• Because when it is multiplied by itself cubically it yields 512.

We subtract it from 571 that is above the 8; 59 remains.

We write the 9 above the 1 and the 6 above the 7.

• We triple the 8; it is 24.

We write the 4 in the [rank] that is third from the 8 that is the sub-triple.

We write the twenty that is indicated by 2 beneath the 8 that we wrote below at first.

We shift the 8 one rank backwards and write it.

There are three digits below: 2 beneath the upper 9; 8 beneath the 7; and 4 beneath the 8.

We write also after the 4 a new digit that is called the present digit, such that when it is multiplied by the three bottom [digits] and by itself cubically, it yields a number that is equal to all the upper digits, or as close as possible. No number that is more suitable for this than 3 is found, so we write 3 beneath the 7 in the first rank.

The multiplication procedure is as follows:

• We multiply the 3 that we wrote by the digit that is worth 20, which is 2; the result is 6. We multiply this 6 by the triple, which is 8; the result is 48.

We subtract the 48 from the digits that are above the 2, which are 59; 11 remains.

• We multiply the 3 for the second time by 4, which is the digit of the units of the first triple above the 8; the result is 12. We multiply it by 8; the result is 96.

• We multiply the mentioned 3 by the mentioned 2; the result is 6. We multiply this 6 again by 3; it is 18.

This 18 with the 96 is 114. We subtract the 114 from the upper digits that are above the 8, which are 117. When we subtract 114 from them, 3 remains.

We write 3 above the 7 and we write zero above each of the ones.

• We multiply the 3 for the third time by 4, which is the units of the triple; it yields 12. We multiply it by 3 itself; the result is 36.

We subtract it from the digits that are above the 4, which are 38; 2 remains.

We write it above the 8; and we write zero above the 3.

• We multiply the 3 for the fourth time by itself cubically; the result is 27.

We subtract it from the digits that are above the 3, which are 27; nothing remains.

Therefore, nothing remains after the extraction of this root, so it is deduced that this number is a perfect cube and its root is 83.

• In order to expand the explanation, we give another example and here is its diagram:

0               2             1 6 0         1 6 4 1 0 0   0 4 2 7 5 5 6 0 1 2 8 1 2 9 0 4

2 2 6 3 3 9 4       2 6

According to what we have said, there are three ranks in this number that have cube roots: the first is the rank of the units, the second is the rank of the thousands, and the third is the rank of the thousands of thousands.

Since one should always begin from the last rank, we start from the 2 that is in the rank of the thousands of thousands and it is the last rank.

We write 2 beneath it, because no other number is found, such that when it is multiplied by itself cubically it closer than 2 to 12 that is above the 2.

• So, we multiply the 2 cubically; it yields 8.

We subtract it from 12; 4 remains.

We write it above the 2 that is in last rank of the upper line; and we write zero above the 1.

• Then, we triple the 2; it is 6.

We write it beneath the digit that is third from the 2 that is beneath the upper 1.

We shift the bottom 2 one rank backwards as stated above, and write the 2 first at the bottom.

Then, we write another digit above the 6 that is the tripled digit, such that when it is multiplied by 6 that is the triple and by 2 that is the sub-triple and by itself cubically, it yields a number that is equal to all the upper digits that are above them, or as close as possible. Since no number closer than 3 is found, we write 3 above the 6.

We multiply in this way:

• We multiply the 3 by the first triple, which is 6; it is 18. We multiply it by the 2 that is next to it, the sub-triple; it is 36.

We subtract it from the digits that are above the 2, which are 48; 12 remains.

We write 2 above the 8; and 1 above the 4.

• We multiply the 3 again by 6; it is 18. We multiply it by 3 itself, which is the present digit; it is 54.

We subtract it from the digits that are above the 6, which are 121; 67 remains.

We write 7 above the 1; 6 above the 2; and 0 above the 1.

• We multiply the 3 again by itself cubically; it yields 27.

We subtract it from the digits that are above it; 5 remains above the 2 and 4 above the 7.

• Then, we triple the 3; it is 9.

We write it in the third rank from it, which is beneath the 0.

We shift the triple, which is 3, and write it in the next rank, so it is placed beneath the upper 9.

We shift the 6 also, and write it [next], so it is placed beneath the upper 5.

We shift the 2 also, and write it in the next rank, so it is placed beneath the upper 4.