Difference between revisions of "עיר סיחון"

Introduction

Since I have seen that arithmetic is a magnificent and extremely lovable science and the Creator, blessed be He, has placed the secret of His unity in the property of number [lit. calculation],

Table of Contents

• The first chapter: on the foundation of all numbers and the nature of the one.

• The second chapter: on the ranks of the numbers and the values of the digits used by the experts of this science.

• The third chapter: on the addition of numerous numbers to each other.

• The fourth chapter: on the subtraction of a small number from a greater number.

• The fifth chapter: on the multiplication of the units by each other, or by others, or on the multiplication of any number by itself, or by another number.

• The sixth chapter: on the division of a great number by a smaller number. In this chapter I will also discuss the scales of multiplication, division, addition and subtraction.

• The seventh chapter: on the extraction of the approximate root of the integer.

• The eighth chapter: on the ratio of a number to another number.

• The ninth chapter: on knowing the fractions of integers - their multiplication, division, addition, or subtraction.

• The tenth chapter: on knowing the factors of square [numbers] that are inexpressible numbers by approximation.

• The eleventh chapter: on nice rules, which are the anecdotes of this science. I will note in this chapter some profound difficult questions with the help of the God of Israel.

Chapter One – The Nature of the One and the Foundation of Numbers

Numeration

• It is a known, clear and true thing that the one is not affected by multiplicity nor by change, but is the cause of any change and multiplicity.

• It is eternal and the origin of all numbers, by which all numbers are generated, for it is their root and foundation and all numbers are growing from it.

Thereof every learned person may understand some hints regarding the secret of the unity of God the Glorious and Awful.

• When we add one to one, which is the first combination, we call it "two".

• When we add one to the two, we call it "three".

• When we add one to the three, we call it "four".

• By adding one to the four we call it "five".

• By adding one to five it is called "six".

• By adding one to six it is called "seven".

• By adding one to seven it is called "eight".

• By adding one to eight it is called "nine".

• Ten in the rank of tens is like one in the rank of units.

• When we have two tens, we call them "twenty".

• To three tens we call "thirty".

• To four tens - "forty".

• To five tens - "fifty".

• To six - "sixty".

• To seven - "seventy".

• To eight - "eighty".

• To nine - "Ninety".

Ninety in the tens is like nine in the units.

When we have ten tens we call them by an inclusive name a hundred, which is the beginning of the [rank of] hundreds, as one is the beginning of the units and ten is the beginning of the tens.

When ten hundred are summed we call them by an inclusive name a thousand, which is the beginning of the thousands, as the hundred, which is the beginning of the hundreds.

• Those are the nine names of the units;

• The name ten;

• The name hundred;

• And the name thousand.

For, twenty, two hundred and two thousand are like two units, each in its own rank.

Also when we say eleven [in Hebrew: "one-ten"], or twelve [in Hebrew: "two-ten"], these names consist of units and tens.

Chapter Two – the Ranks of the Numbers and the Values of the Digits Used by the Experts of this Science

The Positional Decimal System

יספיק לכתוב כל חשבון קטן או גדול בתשע אותיות שנחשוב כל אחת מהן כמו האחדים שהם עד תשעה

• When we want to write twenty, we write a zero first and after is 2, in the second rank, which indicates two tens.

• When we want [to write] a hundred, we write two zeros, then the digit 1 in the third rank, and this indicates a hundred.

Chapter Three – Addition of Numerous Numbers to Each Other

• We write each number of the addition request one beneath the other, each in a row of its own, each rank beneath its corresponding rank, then we draw a line.

• We start from the rank of units and sum all the units.

• We write the sum under the line, if it is less than ten.

• If [the sum] is tens, we write a digit in the column of the second rank, as the number of the tens of the sum and we write a zero alone under the line.

• If the sum is [units] and tens, we should write [the units] under the line and in the second rank [we write] the number of the tens as is stated.

• We proceed this way and do with the rest of the ranks as we did with the rank of units, writing the sums of the ranks successively by the order beneath the line.

• The [total] sum is what comes out from the sum of the ranks arranged under the line.

• If there is one or two ranks in the lines of the addend numbers, of which the whole column are zeros, zeros should be written correspondingly beneath the line as the number of these ranks.

• We wish to know how much is the sum of nine thousand, two hundred and eight, with three thousand, eight hundred and one.

We write these two numbers in two lines, according to this diagram:

the two lines of the numbers

9 2 0 8  3 8 0 1

the ranks of the sum

1 3 0 0 9

[Illustration of the procedure:]

9208		9208		9208		9208		9208
3801		3801		3801		3801		3801
		9		09		1009		13009

• first rank: 9

The sum of the first rank is smaller than ten, hence we write the result beneath the line, so it is 9.

• second rank: 0

Since there is no digit but zeros in the second rank, we write one zero beneath the line, after the 9 that we wrote for the sum of the digits of the first rank.

• third rank: 0

Since the sum of the third rank is ten, we write 1 beneath the lines of the numbers in the following fourth rank and a zero in the third [rank] beneath the line, next to the zero that we wrote after the 9.

• fourth-fifth ranks: 13

The sum of the fourth rank is thirteen, therefore we write 1 for the ten in the fifth rank; we place the rest, which is three, beneath the line next to the last zero and after the three we write the 1, from which we have formed the fifth rank.

Chapter Four – Subtraction of a Small Number from a Greater Number

• We consider all the digits, in whichever rank they are, as if they are units.

• We start from the digit in the last [= highest] rank of the number that is in the bottom line and subtract it from the corresponding digit of the number that is in the upper line.

• We write the remainder above it, if something remains.

• If there is no [remainder], we subtract it by writing a zero above it.

• Then, we proceed with all the other ranks as the procedure in this rank, moving back from rank to rank until they are complete.

• If there is a digit in one of the ranks of the number in the bottom line [= subtrahend] that is greater than the corresponding digit in the upper line [= minuend], there is necessarily one digit in the number of the upper line that follows the [digit] that is in the rank in which we stand.

We do it in this way: we subtract 1 from the digit that follows this rank and write above it what remains from the digit from which we took the 1. If nothing remains from it, as there was only 1 there, we erase it.

It is known that the 1 that we have is worth ten in relation to the preceding rank, hence, we consider the 1 as ten and add to this ten the number of the digit in the rank of the upper number, on which we stand. Now we have enough reserve to subtract the bottom digit from all this and we write the remainder above the corresponding [upper] digit.

• We proceed like this until we reach the first rank and what remains above the line of the upper number is the remainder of the subtraction.

• Whenever we cannot subtract the bottom digit from the upper [digit], we give it supplement from the next rank by the aforesaid way, even if there are zeros separating between it and the following digit, or if we have to subtract the bottom digit from a zero that corresponds it in the upper line.

We shift back by this way: we subtract 1 from the closest digit and write the remainder above it, if something remains from it, then we place this 1 above the preceding zero, so it becomes ten, we subtract 1 from it and 9 remains there above the zero. We place the 1 above the further preceding zero, so it is ten, we subtract 1 from it and nine remains above the zero. So on, we proceed until reaching the digit or the zero of the rank on which we stand, then we do as the aforementioned rule.

• Example: we wish to subtract ninety-two from five thousand eighty-three.

We write the two lines of the numbers according to this diagram: the greater number in the upper line and the smaller number in the bottom line correspondingly.

4 9 9 1  5 0 8 3 9 2

[Illustration of the procedure:]

		4		49		499		4991
5083		1083		183		3
92		92		92		2

• Since the digit in the last rank of the bottom line is greater that the corresponding digit in the upper line, we do as the aforesaid way: we subtract 1 from the last digit of the upper line, which is 5 that is next to the digit in the upper rank, on which we stand, after the zero, and 4 remains instead of the 5.

• The digit 1 becomes ten when we place it above the zero; we subtract one from the ten and nine remains above the zero.

• We place this one above the rank, on which we stand, which is 8, the total is eighteen. We subtract from it the last digit of the bottom line, which is 9; nine remains. We write it in its position above the 8.

• Then, we subtract the 2 of the bottom line from the corresponding 3 in the upper line and 1 remains. We write it above it.

Chapter Five – Multiplication of Units by Themselves or by Other Units and Multiplication of any Number by Itself or by Other

• As if we say, for example: three times four,

• Three times three

• Also ten times twenty

• Or ten times ten

Multiplication of Units by Units

• Two times one is two

• Three times [one] is three

• Likewise, all the other numbers that are multiplied by one do not change or duplicated.

• Two by two is four.

• Two by three is six.

• Two by four is eight.

• Two by five is ten.

• Two by six is twelve.

• Two by seven is fourteen.

• Two by eight is sixteen.

• Two by nine is eighteen.

• Three by three is nine.

• Three by four is twelve.

• Three by five is fifteen.

• Three by six is eighteen.

• Three by seven is twenty-one.

• Three by eight is twenty-four.

• Three by nine is twenty-seven.

• Four by four is sixteen.

• Four by five is twenty.

• Four by six is twenty-four.

• Four by seven is twenty-eight.

• Four by eight is thirty-two.

• Four by nine is thirty-six.

• Five by five is twenty-five.

• Five by six is thirty.

• Five by seven is thirty-five.

• Five by eight is forty.

• Five by nine is forty-five.

• Six by six is thirty-six.

• Six by seven is forty-two.

• Six by eight is forty-eight.

• Six by nine is fifty-four.

• Seven by seven is forty-nine.

• Seven by eight is fifty-six.

• Seven by nine is sixty-three.

• Eight by eight is sixty-four.

• Eight by nine is seventy-two.

• Nine by nine is eighty-one.

Written Multiplication

• We write the two number that we wish to multiply by each other, line beneath line, rank beneath rank and draw a line under the two lines of the numbers.

• We start from the first digit of the upper line and multiply all the upper digits by all the digits of the bottom line successively.

• We write all the products beneath the line in the appropriate ranks, digit by digit, as required.

• We always count how many ranks there are from the digit we multiply on the upper line to the multiplied digit of the bottom line, including these two digits:

• If the product of a digit by another is equal to tens, we write a digit [that mark] the number of the tens according to the counted number of the ranks.

• If it does not reach ten, we write the product on the rank that is one less than the [counted] number [of the ranks].

• If it is equal to units and tens, we write the number of the tens according to the number of the ranks between the multiplied digits and the units one rank before that.

When all the digits of the bottom line are multiplied by all the digits of the upper line according to the rule, we draw a line beneath the ranks of the [interim] products and we sum up all the digits in each rank, according to the addition procedure as explained above.

The result is the [final] product.

• We wish to multiply nine hundred and two by two hundred and forty-six.

We write them according to this diagram:

the lines of the numbers

9 0 2 2 4 6

the interim products

1 3 5 4 1 2   8 6 4 8

the total product

2 2 1 8 9 2

[Illustration of the procedure:]

902		902		902		902		902
246		246		246		246		246
		12		12		412		5412
				8		8		48

	902		902
	246		246
	35412		135412
	648		8648

• When we multiply the digit 2 by the digit 6 beneath it in the bottom line, it is twelve.

The number of the ranks is two, so we write 2 in the first rank, for the two units, and after it, in the second rank, we write 1, for the ten.

• When we multiply the mentioned digit 2 by the 4 in the bottom line, it is eight.

The number of ranks is three, but since this product does not reach ten, we shift it one rank backwards and we write 8 for it in the second rank beneath the 1.

• When we multiply the mentioned digit by the 2 in the bottom line, it is four.

Since this product does not reach ten, we write 4 for this four received now in the third rank after the 1 that is in the second rank.

By this the multiplication of the digit of the first rank in the upper line by all the digits of the bottom line is complete.

Now, we must also multiply the digit 9, which is the last rank of the upper line by all the digits of the bottom line:

• When we multiply this digit by the first digit of the bottom line, which is 6, the result is fifty-four.

The number of the ranks is four, so we write 5 in the fourth rank, for [the fifty, and for] the four, we write another 4 beneath the 4 in the third rank.

• When we multiply this digit by the 4 in the bottom line, the result is thirty-six.

The number of the ranks is five, so we write 3 in a fifth rank for the thirty, after the 5 that is written in the fourth rank, and for the six, we write 6 back beneath the preceding 5.

• When we multiply the mentioned 9 by the last rank of the bottom line, which is 2, the result is eighteen.

The number of the ranks is six, so we write 1 in a sixth rank for the ten, and for the eight, we write 8 back beneath the preceding 3 that is in the fifth rank.

By this all the digits of the upper line are multiplied by all the digits of the bottom line, none of them is missing.

Now, we begin to sum up all the digits in each rank according to the addition procedure, after we draw a line beneath the [interim] products. We write beneath the line the result of addition:

• It is known that the first rank resulting from this multiplication is 2.

• In the second [rank] it is 9.

• In the third [rank] is 8.

• In the fourth rank: 1 remains, after we write the ten of the [sum of the] products in the following rank.

• In the fifth rank: 2 remains, after we write the ten in the following rank, which is the sixth.

• In the sixth rank: we receive 2, since we add to it the ten we have left from the fifth rank that precedes it.

We find that the result is two hundred and twenty-one thousands, eight hundred and ninety-two.

Chapter Six – Division of a Great Number by a Smaller Number

• The one who wants to do this, writes the line of the greater number first, then he leaves a space of a whole line and write the smaller number in another line, corresponding to the first column of the greater number, each rank beneath to its similar.

• Thereafter, he looks and sees how many times he can subtract the number of the last rank in the line of the smaller number [= the divisor] from the digit on the last rank of the greater number [= the dividend] and he subtracts them from it.

However, he should be wise and see if there is a number left, from which he can subtract all the other ranks in the line of the smaller number that precede the last one as the number of times he subtracts this last rank from the last digit in the line of the greater number. If he sees that there will be no number left that will be enough for it, he does not subtract it so many times, but as he sees that is enough to do what is said.

• As the number of times of subtraction he writes a digit corresponding them in the middle between the two lines of the smaller and greater numbers.

• He counts from the last rank in the line of the greater number backwards according to the number of the ranks in the line of the smaller number and beneath the rank where the count ends he writes the number of times of subtraction.

• If there is anything left of the digit of the greater number after he subtracted from it, he writes the remainder above it.

If nothing is left of it, he writes a zero above its.

• Then, he subtracts all the other digits in the line of the smaller number that precede the digit of the last rank, one by one, from the digits that he finds written in the line of the greater number, as much as the times of subtraction from the last digit.

He should be careful throughout to write what remains after the subtraction in its place

If nothing is left of it, he writes a zero above its.

• If he happens to have a digit in one rank of the line of the smaller number that he can not subtract properly from the digit written in the line of the greater number, but there are still digits after the rank, on which he stands, he shifts to it from digits of those ranks, since it is known that each rank is ten times greater than the preceding rank, as is known in the previous chapters. Hence, he takes from their number as needed and writes what remains in its place and zeros, in places where nothing remains.

• After he completes the subtraction of all the digits in the line of the smaller number from the line of the greater number the same number of times, he looks if what remains in the line of the greater number that is not divided yet is greater than the number of the line of the small number. If it is, he looks again, sees and measures how many times he can subtract the digit in the last rank of the line of the smaller number from the last digit of what is left in the line of the greater number and as the number of times he subtracts it from it so are the times he has to subtract the rest of the ranks in the line of the smaller number from what remains then in the line of the greater number.

Therefore, he should be careful first when looking for the value and evaluation to maintain his practice according to the rule and justice.

• As the number of times he subtract the digits that are in the line of the smaller number from the digits of the greater number, he writes a digit corresponding them in the middle between the two lines of the smaller and greater numbers.

• He starts to count from the line of the greater number, from the digit at which he began to look in the second time, according to the number of the ranks in the line of the smaller number and beneath the rank where the count ends he writes the number of times of subtraction.

He should be careful throughout not to forget to write what remains from the line of the greater number.

Or to write a zero above the empty place, where nothing is left.

• Afterwards, he examine if what is left in the line of the greater number is still greater than the number of the line of the smaller number. If it is so, he returns again and sees as we have done twice.

• He does it as many times until the number of what is left in the line of the greater number is smaller than the number in the line of the smaller number.

• If it happens in the beginning of the examination that the number of the last digit of the ranks in the line of the smaller number is greater than the last digit in the line of the greater number, so that he cannot subtract it from it even once, he should shift this whole digit one rank backwards and it becomes a number of tens in this rank, whereupon it will be enough to subtract the digit of the smaller number as many times from it.

He thinks, looks and measures how many times he can subtract it from there, so that the remainder will be enough to subtract from it the other digits in the line of the smaller number the same number of times.

• He always writes the number of times of subtraction in the middle between the two lines of the smaller and greater numbers as the number of the ranks of the smaller number.

• He starts to count their number back from the digit that is in the rank at which he began to look for the subtraction in the line of the greater number.

• We wish to divide one hundred and twenty-five by eleven.

We write the two lines of the numbers according to this diagram:

We write first the greater number, then we leave a space and write the smaller number in another line correspondingly.

0   0 1 4 1 2 5   1 1   1 1

[Illustration of the procedure:]

						0		0
		0		01		01		014
125		125		125		125		125
		1		1		11		11
11		11		11		11		11

• When we look and take the 1 that is in the last rank of the line of the smaller number and the 1 that is in the last rank of the line of the greater number, we know that we can subtract one from the other only once.

Since the ranks of the line of the smaller number are two, we write 1 for this one time beneath the 2 that is in the line of the greater number, because it is second rank to the digit 1 of this same line, from which we start looking.

As nothing remains from this 1, we write a zero above it.

• Then, we take the 1 that in the first rank of the line of the smaller number that precedes the other, with which we have started, in writing and we subtract it once also from the 2 that is in the line of the greater number; 1 remains above it.

By that we have completed the subtraction of the digits in the line of the smaller number from the line of the greater number once of each equally.

When we look at the remainder in the line of the greater number, we see that it is [fourteen] and this number is more than the number of the line of the digits of the smaller number, which is only eleven.

• Because of this, we take again the last 1 from the ranks in the line of the smaller number and subtract it once from the 1 that remains above the 2 in the line of the greater number. Then we write zero above it.

For this subtraction one time, we write 1 beneath the 5, which is the first rank in the line of the greater number, since it is second to the digit, from which we started looking at the subtraction.

• Then, we subtract the first 1 that is in the line of the smaller number once also from the 5, which is the first rank in the line of the greater number; 4 remains above the 5.

Now it is visible and clear to the eye that what remains in the line of the greater number is much smaller than the number in the line of the smaller number.

The number of the middle line that we have formed from the number of the times of subtraction is eleven.

What remains in the line of the greater number is four.

Therefore, we can say that we have found that the smaller number is eleven times in the greater number plus a remainder of four.

• We wish to divide one hundred by nine.

We write them according to this diagram:

0   0 1 1 1 0 0   1 1     9

[Illustration of the procedure:]

				0
		01		011
100		100		100
		1		11
9		9		9

• When we take the 9, which is the smaller number, we see that we cannot subtract it from the 1 that is in the third rank of the greater number.

Therefore, we shift the 1 to the zero that precedes it in writing, so it equals ten and from this ten we subtract 9; 1 remains above this zero.

We write 1 beneath it, for the one time that we have subtracted the 9 from the ten, as it is the first rank of the beginning of looking at the subtraction and the smaller number has only one rank.

We write a zero above the 1 that we have shifted from its position.

• We look again to subtract the 9, which is the smaller number, from the 1 that is left above the zero in the line of the greater number. We should shift it to the other preceding zero, so it equals ten.

When we subtract from it 9 once, 1 remains above the first zero in the line if the greater number.

For this one time that we have subtracted the 9 from the ten, we write 1 beneath the first zero, as it is first to the looking at the subtraction.

By this the division is complete and we have found that there are eleven times nine in one hundred and one is added to them.

• We wish to divide one hundred by twelve.

We write them according to this diagram:

0   0 2 4 1 0 0     8   1 2

[Illustration of the procedure:]

				0
		02		024
100		100		100
		8		8
12		12		12

• When we take 1, which is the last rank in the line of the smaller number and look to subtract it from the 1 that is in the third rank of the line of the greater number, we will not be able to subtract the 2 that precedes the 1 of the smaller number in writing from the zeros that remain there.

Therefore we need to understand the end of the thing from its beginning.

First of all we shift the 1, which is the upper number, above the zero that precedes it in writing, so it equals ten.

Then, we look how many times we subtract the 1 of the smaller number from this ten:

• If we subtract it nine times, only 1 remains from the whole ten, but then we will not have enough to subtract the 2 that is in the line of the smaller number also nine times from what remains in the line of the greater number, because the remainder is only ten, since the 1 is in the second rank, whereas nine times 2 is eighteen.

Therefore we subtract the 1 from the ten that is above the zero in the second rank of the line of the greater number only eight times and 2 remains above that zero.

We write a zero above the 1 that is in the third rank, since we have shift it to the zero and nothing remains there.

For the eight times of subtraction, we write 8 beneath the first zero, since the ranks of the smaller number are two and their counting starts from the second rank in the line of the greater number, as we have started the looking of this subtraction from there.

• Afterwards we subtract also the 2 of the smaller number eight times from what is left in the line of the greater number and we do it in this way:

• We shift the 2 that we have left in the place of the second zero to the first zero, so it is equal to two tens, which are twenty.

• We subtract sixteen from them, which is the product of the times of the subtraction by 2; 4 remains from them above the first zero.

• We write one zero in the place where the 2 that we have shifted to the first zero was written, since nothing remains there.

The result of the division is that there are eight times twelve in one hundred and an excess of four.

• We wish to divide nine hundred and one by thirty-two.

We write them according to this diagram:

0 0   2 2   3 6 5 9 0 1   2 8   3 2

[Illustration of the procedure:]

						0		00
				2		22		22
		3		36		36		365
901		901		901		901		901
		2		2		28		28
32		32		32		32		32

• We see that if we subtract three times the 3 that is in the bottom line, which is the smaller number, from the 9 that is in the upper line, which is the greater number, we will not be able afterwards to compensate the other digit in the bottom line in order to subtract it from the zero, since nothing remains from the 9.

Therefore we subtract it only twice and we write the remaining 3 above it.

For these two times of subtraction, we write 2 beneath the zero, since the ranks of the bottom line are two and the zero is also second to the digit, from which we have started the subtraction.

• Then we subtract the digit 2, which is the first rank in the bottom line, also twice from the zero that is in the upper line and we do it in this way:

• We take 1 from the 3 that remained above the 9 and 2 remains above it.

• We put the 1 that we took above the zero, so it equals ten. We subtract two times two from it; their product is four and 6 remains above the zero.

Now we have made a compensation by subtracting all the digits of the bottom line from the upper line an equal number of times [for each].

• Indeed, when we look at what is left in the upper line, we see that it is greater than the number of the bottom line, and because of this we should look again to subtract the 3, which is the last rank in the bottom line, from the 2 that is left in the last rank of the upper line. We do it in this way:

• We shift the whole 2 to the 6 above the zero; they are twenty-six.

• Now, we measure how many times we can subtract the 3 from the twenty-six: we can subtract it eight times, which are twenty-four.

• When we subtract them from twenty-six, 2 remains above the 6.

• We write zero above the last rank of the upper line, since nothing remains there.

• For the eight times of subtraction, I write 8 beneath the 1 that is the first rank of the upper line, since it is second to the rank, from which we have started looking at the subtraction.

• Now, we subtract also the digit 2 of the bottom line eight times from what is left in the upper line. We do it in this way:

• We shift the 2 that is left in the second rank of the upper line to the 1 that is in the first rank; they are twenty-one.

• We subtract from them the product of 8 by 2, which is sixteen; 5 remains above the 1, which is the first rank of the upper line.

• We write a zero above the 2 that is written in the second rank, since we have shifted the whole of it to the first rank.

This division is complete and we have found that the middle line is twenty-eight and what remains above the upper line is five.

• We wish to divide eight hundred and ninety-one by forty.

We write them according to this diagram:

0 1   8 9 1   2 2   4 0

[Illustration of the procedure:]

		0		01
891		891		891
		2		22
40		40		40

• We subtract the 4 that is in the bottom line twice from the 8 that is in the upper line.

We write above it a zero, since nothing is left of it.

For the [two] times of subtraction, we write 2 beneath the 9 that is in the upper line, since it is second rank to the digit, from which we have started the subtraction and in the bottom line there are also two ranks.

• Again, we subtract the known 4 twice from the corresponding 9 in the upper line; 1 remains above it.

For the two times of subtraction, we write 2 beneath the 1 that is the first rank of the upper line, since it is second to the digit 9, from which we have now started the subtraction.

The result of the division is that in eight hundred and ninety-one there are twenty-two times forty and an excess of eleven.

• We wish to divide three hundred and twenty-one by nine.

We write them according to this diagram:

0   0 5 6 3 2 1   3 5     9

[Illustration of the procedure:]

				0
		05		056
321		321		321
		3		35
9		9		9

• When we try to subtract the 9, which is the smaller number, from the 3 that is in the last rank of the upper line, we have to shift the whole 3 to the 2 that precedes it in writing; they are thirty-two.

We subtract the 9 from them three times, which are twenty-seven, 5 is left from them above the 2.

For these three times of subtraction, we write 3 beneath the 2 that is in the upper line, since we start the subtraction from this rank and in the bottom line there is only one rank.

We write a zero above the rank of the 3, since nothing is left of it.

• Again, we subtract the 9 from the 5 that is left above the 2 in the upper line, but since we cannot subtract the 9 from the 5, we shift the whole 5 to the 1; they are fifty-one.

We subtract the 9 five times, 6 is left from them above the first rank.

We write a zero above the 5, since nothing is left of it.

For the five times of subtraction, we write 5 beneath the first rank, from which we have started the subtraction, since in the bottom line there is only one rank.

This division is complete and the middle line is thirty-five and what remains above the upper line is six.

• We wish to divide five hundred and eighty-three thousand, six hundred and ninety-six by seven hundred and sixty-four.

0 0       0 3 2       4 6 0 0     5 8 2 1   0 9 1 8 5 0 5 8 3 6 9 6       7 6 4       7 6 4

[Illustration of the procedure:]

				5
		09		091
583696		583696		583696
		7		7
764		764		764

	4
	58
	0918
	583696
	7
	764

The skill of what is required is much greater and deep than all we have mentioned and this is its description:

• When we look to subtract the 7 that is in the bottom line from the 5 that is in the upper line, we cannot do so.

• Therefore, we shift it to the [8], which is the rank that precedes it in writing, and write a zero in its place, since nothing is left there.

• The result is fifty-eight and when we measure how many times we shall subtract the 7 from it, we find that we can subtract only 7 times, so that we will be able to do the necessary rule with the remainder.

• Seven times 7 is forty-nine and when we subtract it from fifty-eight, 9 remains above the 8.

• We start counting three ranks of the bottom line back from this rank and the counting ends at 6 that is the third rank; we write 7 beneath it, for the seven times of subtraction.

• Then, we subtract the 6 that is in the bottom line also 7 times from the 9; the result of the multiplication is forty-two. We do it in this way:

• We subtract 4 from that 9 and write the remainder, which is 5, above it.

• We shift the 4 to the 3, which is the rank that precedes in writing the rank above which the 9 is; they are forty-three. We find that 1 remains from them above the 3.

• Then, we subtract the 4 that is in the bottom line, which is the first rank of it, also 7 times; the result of the multiplication is twenty-eight. We do it in this way:

• We take 1 from the 5 that is left in the upper line, in the fifth rank, and shift it to the 1 that is left above the 3 that precedes this rank; they are eleven. We subtract 3 from them; 8 remains above the 1.

• We shift the 3 to the rank that precedes its rank in writing, which is 6; they are thirty-six. We subtract from them the product of 4 by 7, which is twenty-eight; 8 remains above the 6.

Now, we have completed the subtraction of all the digits of the bottom line from the upper line equally, each as many times as the other.

	0		03
	46		46
	58		582
	0918		0918
	583696		583696
	76		76
	764		764

	03
	460
	582
	09185
	583696
	76
	764

When we examine what is left in the upper line, we see that it is much more than the number of the bottom line.

• Therefore we look again to subtract the 7 that is in the bottom line from the 4 that is left in the upper line, in the fifth rank.

• We have to shift it to the 8 that is left above the preceding rank; they are forty-eight.

• When we measure, we find that we can subtract the 7 six times from them, the product of which is forty-two, and when we subtract it from them, 6 remains above the 8.

• We write a zero above the 4, since nothing is left of it.

• We start counting from the rank, above which we have wrote the 6, back by the three ranks of the bottom line, ending at the 9, which is the second rank of the upper line, and there we write 1 beneath it, for the six times of subtraction.

• Then we have to subtract the 6 that in the bottom line also six times; their product is thirty-six. We do it in this way:

• We subtract three from the 6 that is left in the fourth rank; 3 remains above the 6.

• We shift the 3 that we have subtracted to the 8 that is left in the rank that precedes this rank; they are thirty-eight. When we subtract thirty-six from them, 2 remains above the 8.

• We have to further subtract the 4 that is in the bottom line also six times; their product is twenty-four. We do it in this way:

• We take the 2 that is left above the 8 and shift it to the 9 that is the rank that precedes this rank in writing; they are twenty-nine. We subtract twenty-four from them; 5 remains above the 9.

• We write a zero above the rank, in which the [2] is written, since nothing is left of it.

By this, we have completed again the subtraction of the digits of the bottom line from the upper line equally, each the same number of times.

	0		00
	032		032
	460		460
	582		5821
	09185		09185
	583696		583696
	764		764
	764		764

	00
	032
	4600
	5821
	091850
	583696
	764
	764

When we examine what is left in the upper line, we find that it is still more than the number of the bottom line.

• Therefore we should look again to subtract the 7 that is the last rank of the bottom line from the 3 that is left in the fourth rank.

• It is necessary to shift this 3 to the third rank that precedes it in writing, in which a zero is written; so the 3 equals thirty.

• We subtract the 7 4 times, for it will be enough for us to apply the necessary rule upon the remainder.

• We write a zero above the 3, since nothing is left there.

• When we subtract twenty-eight from thirty, 2 remains above the zero that is in the third rank.

• We count from this rank back by the three ranks of the bottom line, ending at the first rank; we write beneath it the 4 times of subtraction.

• Then we have to subtract the 6 that in the bottom line also 4 times; their product is twenty-four. We do it in this way:

• We shift the 2 that is left in the third rank to the 5 that is left in the rank that precedes it in writing; they are twenty-five.

• When we subtract twenty-four from them, 1 remains instead of the 5.

• Above the 2 we write a zero, since nothing is left of it.

• We have to further subtract the 4 that is the first rank of the bottom line also 4 times; their product is sixteen. We do it in this way:

• We take the 1 that is left in the second rank, write above it a zero and shift it to the 6 that precedes this rank; they are sixteen. We subtract the product of 4 by 4 from them.

• We write a zero above the 6, since nothing is left there.

We find that the upper line is all gone in the divisions we made and the middle line that is created is the same as the line of the smaller number, no more and no less.

• We wish to divide five hundred and eighty-three thousand, six hundred and ninety-six by one thousand and eighty.

We write them according to this diagram:

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

[Illustration of the procedure:]

		0		04
583696		583696		583696
		5		5
1080		1080		1080

	0		0
	04		0404
	583696		583696
	54		540
	1080		1080

• We see that we are able to subtract the 1 that is in the bottom line five times from the 5 that is in the upper line.

• We do as we said and we write above it a zero, since nothing is left of it.

• We count four ranks back from this rank that is in the upper line, for the four ranks that are in the bottom line; their count ends in 6, which is in the same line. We write there 5, beneath it, for the five times of subtraction.

• Then, we have to subtract the 8 that is in the bottom line also 5 times; their product is forty. We do it in this way:

• We subtract 4 from the 8 that is in the upper line and shift the 4 to the rank that precedes the 8 in writing, which is 3; they are forty-three. The forty are all gone and the 3 remains at place.

By this, all the digits of the bottom line were subtracted from the upper line equally, each the same number of times.

When we examine what is left in the upper line, we see that its number is more than the number of the bottom line.

• Therefore we look again to subtract the 1 that is in the bottom line 4 times from the 4 that is left in the fifth rank.

• It is all gone, so we write a zero in its place.

• Count four ranks back; it ends in the second rank. We write there 4, beneath it, for the four times of subtraction.

• We have to further subtract the 8 of the bottom line also 4 times; their product is thirty-two. We do it as follows:

• We take the 3 that is left in its place, in the fourth rank of the upper line, and shift it to the digit of the third rank, which is 6 that precedes it in writing; they are thirty-six. When we subtract thirty-two from them, 4 remains above the 6.

• I write a zero above the 3, since nothing is left from it.

By this, all the digits of the bottom line were subtracted from the upper line again.

When we examine what is left in the upper line, we see that its number is less than the number of the bottom line, and this is your sign that the division is complete.

The middle line generated from the number of the times of subtraction is five hundred and forty.

The remainder above the digits of the upper line is four hundred and ninety-six.

Methods of Checking - Multiplication, Division, Addition, Subtraction

• Checking the result of division: multiplication

I say that the most truthful and right scales are that whoever does the division operation multiplies the middle line that is generated from the number of the times of subtraction, according to the procedure of the multiplication operation, by the bottom line, which is the smaller number; then he takes all that remains above the upper line, if something remains there; adds it beneath the product, each digit beneath its corresponding rank; and sums it with the whole product. The line of the sum must be the same as the line of the greater number that he divided, each digit the same as its corresponding digit, if his deed is pure or upright [Proverbs 20, 11].

• Checking the result of multiplication: division

Likewise, the scales of the multiplication operation are that he divides the result of multiplication by one of the two lines of the multiplied numbers and the result of division is the second line, if he goes the right way.

• Checking the result of addition: subtraction

Also, whoever adds two lines of numbers to each other, uses the subtraction operation: he subtracts one of the two lines from the sum and if his procedure is correct, he is left with the second line.

• Checking the result of subtraction: addition

According to our ways, we understand and learn that whoever subtracts a small number from a greater number, if he adds what is left for him to the line of the small number, the sum he receives is inevitably the same as the line of the greater number.

Chapter Seven – Extracting the Closest Root of Integer

One versus the integers

• For all integers, you find that when you multiply a number by itself, it, meaning the product, becomes greater than the original number.

Except for one, because when you multiply one by itself it is not affected by any change, but the result is one as it was originally without any multiplicity or change.

Therefore we can say that the number one is a square as well as the root of itself.

• The number one has another advantage over all other numbers:

For all other numbers, when we take their extremes, [i.e.] the number that precedes them and the number that follows them, the sum of the two extremes is twice the mean number.

But, one does not have a preceding extreme. Its following extreme, which is two, is itself its double.

Square Nnumbers

• I say regarding the roots that if you multiply a square number by a square number, the product is also a square number.

When you want to know its root, you only need to extract the roots of the numbers you multiply by each other. Multiply the roots by each other and the result is the root of the third number.

• Example: four is a square number and nine also. The product of the one by the other is thirty-six and it is also a square number. If we want to know its root, we extract the root of four, which is two, and the root of nine, which is three; we multiply them by each other; the result is six and this is the required. Because, when we multiply six by itself, it is thirty-six.

• By the root of the ranks, I mean that when you look at the first rank, which is the rank of units: its first number, which is one, has a root, as [explained] above.

• The second rank, which is the rank of tens: its first number, which is ten, has no root.

• Also the third rank: its first number, which is a hundred, has a root.

• The fourth rank: its first number, which is a thousand, has no root.

All the ranks follow this way.

Extracting roots - written procedure

• We wish to know the root of two hundred and twenty-five.

We write it in the following diagram:

0     1 0 0 2 2 5   2 5

[Illustration of the procedure:]

				0		0
		1		100		100
225		225		225		225
		12		25		15

• Then, we count the number of the ranks and we see that the last [rank] has a root, since it is third, therefore we start with it.

• It is 2. We look for the approximate root of 2; it is one, so write it beneath this 2.

• We subtract its product by itself from it; 1 remains above it.

• We multiply twice the 1 that we wrote beneath the 2, and write the product back beneath the preceding rank.

• So, we write 2 beneath the 2 of the second rank, and we write a zero above the 1 we wrote beneath the 2 of the last rank, or cross it out by a pen, as a sign that it is erased from there as if it were not written there.

• Then, we shift the remaining 1 in the first rank from which we began, above the 2 that precedes it in writing; it is twelve.

• We examine how many times we can subtract the 2 from it, so that we will be able to subtract the product of the times of subtraction by themselves from the number that remains from it.

• We find that if we subtract it five times it will be enough for the required and 2 remains in its place.

• We write a zero above the 1, as nothing is left of it.

• We write 5 beneath the first rank for the five times of subtraction.

• Then we multiply the times of subtraction by themselves; the result is twenty-five.

• We subtract it from the twenty-five that is in the line of the examined number and we find that it is all gone.

• So, we write zero above the 5 and above the 2.

• We create another line beneath it, which is 25.

• We subtract from 2 its half; 15 remains and this is the root of the sought number.

For, if you multiply 15 by itself, you find the result is 225.

• We wish to know the approximate root of nine hundred and twenty-five.

This is its diagram:

0     9 2 5   6 0

[Illustration of the procedure:]

		0		0		0
925		925		925		925
		36		60		30

• Its last rank has a root, therefore we start from it.

• Its root is 3; we write it beneath it.

• We subtract the 3 it; it is all gone, and nothing is left of it. Therefore we write a zero above it.

• We multiply the 3 twice, cross it out with a pen and write 6 one rank back.

• We get the approximate root, because now we cannot subtract further 6 from 2 that is above it. So, we write a zero beneath the first rank, and we find that the line created is 60.

• We subtract from the 6 its half; 30 remains and it is the required approximate root.

For, if you multiply 30 by itself and put the remainder, which is 25, beneath it in the first line, you get 925.

• We wish to know the root of seven thousand and fifty-six.

We write it in the following diagram:

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

[Illustration of the procedure:]

				0		0
				20		20
		06		0610		0610
7056		7056		7056		7056
		86		164		164
		1				8

• The last rank of this number has no root, since it is fourth, therefore we shift it to the zero that precedes it in writing, so it is seventy.

• The approximate root of seventy is 8, whose product by itself is sixty-four.

• We subtract it from seventy; 6 remains above the zero.

• We write a zero above the last rank, as nothing is left of it.

• For the approximation root of seventy, we write 8 beneath the zero that is in the third rank.

• Then we multiply the 8 twice; it is sixteen. We write it this way:

• 6 for the six beneath the rank that precedes the zero.

• 1 for the ten beneath the 8, which we cross it out with a pen.

• Afterwards we examine how many times we can subtract the 1 and the 6 from the 6 and the 5 that are above them, so that enough is left of them for the product of the times of subtraction by themselves.

• We find that we can subtract 4 times the 1 from the 6 that above it and 2 remains above it.

• Then we subtract the 6 also 4 times that are twenty-four, when we shift the 2 to the 5 that precedes it, so they are twenty-five and 1 remains above the 5.

• We write a zero above the 2, as nothing is left of it.

• For the four times of subtraction we write 4 beneath the first rank.

• Then, we multiply the 4 by itself; the result is sixteen.

• We subtract it from the sixteen that is in the line of the examined number and we find that it is all gone.

• So, we write zeros above the 6 and above the 1.

• The line that is generated is 164.

• We take half the 6 and the 1 this way:

• We shift 1 above the 6; it is sixteen and its half is 8.

• We write it be neath the 6.

• We find that the remainder is 84 and this is the root of the sought number.

For, if you multiply it by itself, you get one line, whose number is seven thousand and fifty-six.

• We wish to know the the approximate root of seventy-six thousand, five hundred and forty-three.

This is its diagram:

0 3       3 8 6     7 6 5 4 3   4         1 4

[Illustration of the procedure:]

				03
		3		386
76543		76543		76543
		24		47
				14

• The last rank has a root, since it is fifth, so we start from it.

• The approximate root of 7 is 2; we write it beneath it.

• We subtract from it the product of 2 by itself; 3 remains above the 7.

• We multiply the 2 twice, cross it out with a pen and write the product, which is 4, one rank back, beneath the 6 that is the fourth rank.

• Then, we shift the 3 that remains above the [7] to the rank that precedes it in writing; it is thirty-six.

• We examine how many times we can subtract from it the 4 that we wrote beneath the four rank, so that enough is left of it for the product of the times of subtraction by themselves.

• We find that we can subtract it only seven times that are twenty-eight.

• We subtract it from thirty-six; 8 remains above the 6.

• We write a zero above the 3 that was shifted from its place.

• For the seven times of subtraction we write 7 beneath the third rank.

• Then, we multiply 7 by itself; the result is forty-nine.

• We subtract it from what remains in the line of the number this way:

• We take 5 from the 8 that remains above the fourth rank; 3 remains above it.

• We shift the 5 to the 5 that precedes it in writing, which is the third rank; it is fifty-five.

• We subtract forty-nine from it; 6 remains above the 5.

• Then, we multiply the seven times of subtraction twice; it is fourteen.

• We cross the 7 out with a pen.

• We write this way:

• 4 for the four beneath the 7 that we crossed out with a pen.

• 1 for the ten beneath the 4 that is the fourth rank.

• Thereafter, we write what remains in the line of the examined number in another place and make it a separate line; then we write beneath it what is obtained.

The line of what remains is 3643.

We write beneath it in [another] line: 4 beneath the 4 that is in the second rank, and 5 beneath the 6; all this is what was obtained above.

It is written in this form:

3 6   0 4 0 7 3 6 4 3   5 4 6

[Illustration of the procedure:]

	36		36
	0407		0407
	3643		3643
	546		276

• Now, we shift the 3 that is in the last rank of the upper line to the 6 that precedes it; it is thirty-six.

• We write zero above the 3 that was shifted from its place.

• We examine how many times we can subtract the digits of the bottom line from the upper line, so that we will be able to subtract the product of the times of subtraction by themselves from the number that remains of it.

• We see that it is enough for that if we subtract it only 6 times.

• When we subtract the 5 six times that are thirty from the thirty-six, the 6 stays in its place.

• We subtract also the 4 that is written beneath the second rank six times; the result is twenty-four.

• We do it this way:

• We take 2 from the 6; 4 remains above it.

• We shift the 2 to the 4 that precedes this rank in writing; the result is twenty-four that are all gone.

• We write zero above the 4, since nothing remains there.

• For the six times of subtraction we write 6 beneath the first rank.

• We multiply the 6 by itself; the result is thirty-six.

• We subtract it from what is left in the upper line this way:

• We subtract 1 from the 4 that is left above the third rank; 3 remains there.

• We shift the 1 to the zero, which is above the second rank that precedes it; it is ten.

• We subtract four from it; 6 remains above the zero.

• We shift the 4 to the first rank; it is forty-three.

• We subtract thirty-six from it; 7 remains above the 3 that is in the first rank.

• The bottom line that is generated is 546.

• We take half the 4 and the 5; 27 remains.

• We find that the remainder after all this is 276 and this is the approximate root of the sought number.

If you multiply 276 by itself and add what remains in the upper line, which is 365, you get the line of the number whose root you sought.

• We wish to know the root of five hundred and eighty-three thousand, six hundred and ninety-six.

We write it in the following diagram:

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

[Illustration of the procedure:]

				0
				36
		09		0990
583696		583696		583696
		74		746
		1		112

• Since the last rank of this number has no root, the 5 should be shifted to the 8 that precedes it in writing in the fifth rank; it is fifty-eight.

• The approximate root of this number [is 7], whose product by itself is forty-nine.

• When we subtract it from fifty-eight, 9 remains above the 8.

• We write zero above the 5 that was shifted from its place.

• For the approximate root we write 7 beneath the 8.

• [We] multiply the 7 twice, and cross it out with a pen; the product is fourteen.

• We write four for the four beneath the 3 that is the fourth rank and 1 for the ten beneath the 7 that we crossed out with a pen.

• Now, we should [examine] how many times we can subtract the 1 and the 4 from what remains in the upper line, so that we will have enough to subtract the product of the times of subtraction by themselves from the remainder.

• We find that it is enough for us if we subtract it six times.

• When we subtract the 1 six times from the 9, 3 remains above it.

• We also have to subtract the 4 six times; the product is twenty-four.

• We do it this way:

• We take the 3 that remains above the 9 and write a zero in its place.

• We shift it to the 3 that precedes it in the fourth rank; the result is thirty-three.

• When we subtract the twenty-four from it, 9 remains above the 3.

• For the six times of subtraction we write 6 beneath the third rank.

• When we multiply it by itself the result is thirty-six.

• We subtract it from the line of the number this way:

• We take 3 from the 9 that is left in the fourth rank; 6 remains above it.

• We shift the 3 to the 6 that precedes it in the third rank; it is thirty-six, which is a product of the 6 by itself.

• We write a zero above the 6, for no remainder is left.

• Then, we double the six times of subtraction; it is twelve.

• We write it this way:

• 2 beneath the 6 for the two and we cross it out by a pen.

• 1 beneath the 4 that follows for the ten.

• After all this, we write what remains in the line of the number in another place and make it a separate line; then we write beneath it what is obtained.

The line of what remains is 6096.

We write beneath it in another line: 2 beneath the 9, 5 beneath the zero, and 1 beneath the 6; all this is what was obtained above.

It is written in this form:

0       2   1 0 6 0 9 6 1 5 2 4

[Illustration of the procedure:]

	0 0		36
	2 10		2 10
	6096		6096
	1524		764

• Now, we should examine how many times we can subtract the digits that we wrote beneath the first line, so that we will have enough afterwards to subtract the product of the times of subtraction by themselves from the remainder.

• We find that it is enough for all this if we subtract it 4 times.

• When we subtract four times the 1 that is the last rank of the bottom line from the 6 that is in the upper line above it, 2 remains above it.

• We have to subtract from the upper line also four times the 5 that is in the bottom line, the product of which is twenty.

• We do it this way:

• We take the 2 the is left in the last rank of the upper line and shift it to the zero that precedes that rank; it is twenty and all is gone for the twenty of the product.

• We write a zero above the 2 that we took, as it was shifted from its place.

• Then, [we] subtract also four times the 2 that is in the bottom line from the 9 that is above it; 1 remains above it.

• For the four times of subtraction we write 4 beneath the first rank of the upper line.

• When we multiply 4 by itself, it is sixteen.

• We subtract it from the sixteen that is in the upper line, because 1 is left above the 9 in the second rank and the 6 in the first [rank].

• We find that the whole upper line is gone, so we write zeros above the 6 and the 1 that above the 9.

• The bottom line that is generated is 1524.

• We take half the 2, 5 and 1; the remainder is 764 and this is the sought root of the number.

If you multiply 764 by itself, I promise you that the product is 583696.

• We wish to know the root of this number, which is eight hundred and twenty-four thousand, four hundred and sixty-four.

We write it in the following diagram:

0 0       0 1 6 0 0 0 8 2 4 4 6 4   9 1 8 0 8

[Illustration of the procedure:]

				00		00
		01		016000		016000
824464		824464		824464		824464
		9180		91808		908

• Since the last rank of this number has no root, it should be shifted to the 2 that precedes it in writing that is the fifth ranke; it is eighty-two.

• Its approximate root is nine, since its product by itself is eighty-one.

• When we subtract it from it, 1 remains above the 2.

• We write a zero above the 2, as it was shifted from its place.

• We write 9 beneath that 2 for the nine, which is the approximate root and cross out the 9 with a pen.

• We double it; it is eighteen.

• We write it this way:

• 1 beneath the fourth rank, for the ten.

• 8 for the eight one rank back.

• Then, we write a zero beneath the second rank.

• Then we examine how many times we can subtract the digits of the created bottom line from what remains in the upper line, so that a number remains there from which we will be able to subtract the product of the times of subtraction by themselves.

• We find that it is enough to subtract it eight times.

• We do it this way:

• We take the 1 that is left in the upper line above the fifth rank and write zero above it, since we shift it from its place.

• We shift it to the 4 that is left in the fourth rank that precedes it; it is fourteen.

• We subtract from it eight times the 1 that beneath the fourth rank; 6 remains above the 4.

• We also have to subtract eight times the 8 that is beneath the third rank; its product is sixty-four.

• We do it this way:

We take the 6 left in the fourth rank and write a zero above it when we shift it from its place.

We write it above the 4 in the third rank that precedes it; it is sixty-four and it is all gone when subtracting the product of 8.

We write a zero above the 4, because nothing remains of it.

For the eight times of subtraction we write 8 beneath the first rank.

• When we multiply 8 by itself the result is sixty-four.

We subtract it from the sixty-four above it in the line of the number examined, because so far the 4 remains in the the first rank and the 6 in the second rank. Thus the entire upper line is gone.

So, we write zeros above the 4 and the 5.

• The bottom line that is generated is 1808.

A half of 180 is 90. We find that what remains is 908 and this is the required root of the number.

Check: Try me herewith [Malachi 3, 10]: to multiply 908 by itself, for you will necessarily receive 824464 from the multiplication.

• We also want the root of a hundred.

We write it in the following diagram:

0     1 0 0   1 0

[Illustration of the procedure:]

		0
100		100
		1

The third rank has a root, so we start with it.

The root of 1 is one.

So, we write a zero above the 1, because it is gone when we subtract from it the product of the root by itself.

Since it is the root, we write 1 beneath the second zero. It indicates a ten there and this is the required root of the number.

• We wish to know the root of four hundred.

We write it in the following diagram:

0     4 0 0   2 0

[Illustration of the procedure:]

		0
400		400
		2

It is known that the root of 4 is two.

Therefore, we write a zero above the 4, because it is all gone, when you subtract from it the product of the root by itself.

For the two, which is the root, we write 2 beneath the second zero. It indicates twenty there, and this is the root of the number.

• If we wish to know the approximate root of four thousand.

[Illustration of the procedure:]

				0		0
				13		13
		04		0441		0441
4000		4000		4000		4000
		62		623		63
		1		1

We do this way:

The fourth rank has no root, so we write 4 above the zero that precedes it; it becomes forty.

The closest [root] is six, because its product by itself is thirty-six.

4 remains above the third zero.

We write a zero above the 4 that was in the fourth rank, when it was shifted from its place.

For the six, which is the closest root, we write 6 beneath the third zero.

Then, we multiply 6 twice and cross it out with a pen; it becomes twelve.

We write 2 beneath the second zero for the two and 1 for the ten beneath the 6 [that was erased] with a pen.

When we examine how many times we can subtract these 2 and 1 from what remains ubove the line, we find that three times are enough for us.

We do this way:

We subtract the 1 beneath the third zero three times from the 4 remaining above it; 1 remains instead of it.

We also have to subtract [three] times the 2, the product of which is six.

We do it this way:

We take the 1 remaining in the third rank and shift it above the second zero; it becomes ten.

We subtract the six from it; four remains there.

We write zero above the 1 that is above the third rank, because it was shifted from its place.

For the three times of subtraction we write 3 beneath the zero that is in the first rank.

When we multiply it by itself, it is nine.

We subtract it from the upper line this way:

We take 1 from the 4 remaining in the second rank; 3 remains above it.

We shift the 1 to the zero that precedes its rank; it becomes ten.

We subtract nine from it; 1 remains above that zero.

The bottom line that is generated is 123.

A half of 12 is 6.

We find that the required approximate root [is 63]

Chapter Eight – the Ratio of a Number to Another Number