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],

ומחשבתי הטובה יצרף אותה למעשה
למען רחמיו וחסדיו אשר נפלאו ועצמו מספר כל מעמדיו
יהי שמו משובח לעד ומפואר לנצח

Numbers 21, 27: בֹּאוּ חֶשְׁבּוֹן; תִּבָּנֶה וְתִכּוֹנֵן, עִיר סִיחוֹן = "Come to Ḥeshbon, may it be built and established as the city of Siḥon"

Table of Contents

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

Numeration

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

• Two

• Three

• Four

• Five

• Six

• Seven

• Nine

• Ten

ten in the rank of tens is like one in the rank of units

• Twenty

• Thirty

• Forty

• Fifty

• Sixty

• Seventy

• Eighty

• Ninety

all the names of the tens are borrowed from the units

When we have ten tens we designate 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 with an inclusive name a thousand, which is the beginning of the thousands, as 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.

Chapter Two – the Decimal Ranks and the Ten Digits

The Positional Decimal System

יספיק לכתוב כל חשבון קטן או גדול בתשעה אותיות שנחשוב כל אחד מהם כמו האחדים שהם עד תשעה

• twenty: 20

• hundred: 100

Chapter Three – Addition

• We write each number of the addition request one beneath the other, each in a row of its own, rank by 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

• 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 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 another 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 one 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-nine 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

• Since the digit in the [highest] 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 becomes ten when we place it above the zero; we subtract one from the ten and nine remains.

• 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

• For example: three times four

• When we say three times three

• We say 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 eight 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 bottom digits 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.

Summing the interim multiples

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

• first rank: 2

• second rank: 9

• third rank: 8

• fourth rank: 1

• fifth rank: 2

• sixth rank: 2

The result: 221892

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 corresponds to its similar.

• Thereafter, he looks and sees how many times he can subtract the number of the digit on 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 from the subtraction of the smaller 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 assists 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, if that which is not divided yet is greater than the number of the line of the small number, 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 we 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 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

• 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 clearly 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 similar 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 zero that is in the first rank.

• 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

• 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 3 that remains from the 9 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, we 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, five 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, which is the first rank preceding in writing; they are fifty-one.

We subtract the 9 from them 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

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.

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.

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 digit 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

• 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.

• We 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; the product of 4 by 8 is thirty-two. We do it in this way:

• 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.

• We 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 the sign that the division is complete.

The middle line generated from the number of the times of subtraction of the digits in the ranks 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 checking method is 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 checking method of the multiplication operation is 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

• Checking the result of subtraction: addition

Chapter Seven – Extracting the Closest Root of Integer

One versus the integers

• For all integers - the product of an integer by itself is greater than the integer itself

Except for one - one is not affected or changing when multiplied by itself

Therefore it is a square as well as the root of itself

• Another advantage of one over all other numbers

For all other numbers - the sum of the preceding number and the succeeding number is equal to twice the middle number

One does not have a preceding extreme, only subsequent extreme - its single extreme is equal to its double

square numbers

• The product of a square by a square is a square

• Square ranks and non-square ranks

• The first number of the units - the number one - is a square number

• The first number of the tens - the number ten - is not a square number

• The first number of the hundreds - the number one hundred - is a square number

• The first number of the thousands - the number one thousand - is not a square number

And so on

Extracting roots - written procedure

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

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 closest root, because now we cannot subtract further 6 from 2 that is above it, and since it is beneath the first rank, 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.

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 root [whose square] is the closest to 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 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 the 1 4 times 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.

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

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

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

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

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

0     1 0 0   1 0

[Illustration of the procedure:]

		0
100		100
		1

0     4 0 0   2 0

[Illustration of the procedure:]

		0
400		400
		2

Chapter Eight – Proportions

1) Rule of Three

2) Rule of Four

3) Arithmetic proportion

• 1; 2; 3; 4

• 2; 4; 6

This kind is clear and explained - thus, there is no need to elaborate on that

4) Harmonic proportion

Chapter Nine – Fractions

Introduction

Operations with fractions

Multiplication of fractions

Introduction

the common denominator of the fractions is

Multiplication of fractions by fractions

• We wish to multiply four fifths by four fifths

• It is known that the fifth is derived from the five

• We wish to multiply two quarters by three quarters

• We wish to multiply three quarters by four fifths.

Check:

Multiplication of integers by fractions

Multiplication of integers and fractions by fractions of one type

נוסיף עליהם הארבע חמשיות ויהיו תשעה עשר חמשיות
נכפול אותם על השלש חמשיות יהיו חמשים ושבע חמשיות חמשית
והחמשים מהם הם שני שלמים והשבע חמשיות חמשית הנשארות הן הם חמשית אחת ושתי חמשיות חמשיות

Multiplication of integers and fractions by fractions of different types

Multiplication of integers and fractions by integers and fractions of one type

Multiplication of integers and fractions by integers and fractions of different types

Division of fractions

Division of fractions by fractions

Division of integers and fractions by integers and fractions

Addition of fractions

Subtraction of fractions

Chapter Ten – Approximating the Roots of Non-Square Numbers

Introduction: square numbers and non-square numbers

Sexagesimal fractions

• The astronomers divided the zodiac into 12 zodiacal signs

The reason for dividing by 12: twelve is the smallest number that has numerous fractions [= divisors, in modern terminology]:

• Each zodiacal sign was divided into 30 degrees

The reason for dividing by 30: thirty is the smallest number that has as many fractions [= divisors, in modern terminology] as it has:

• Each degree was divided into 60 minutes

The fractions [= divisors] of 60:

• Each minute was divided into 60 seconds

• Each second was divided into 60 thirds

• Each third was divided into 60 fourths

Multiplication of Sexagesimal Fractions

Written procedure

Extracting Roots of Deaf Numbers

1) First approximation:

2) Second approximation:

3) Third approximation - repeating the procedure of the second approximation - the most accurate approximation

1) First approximation:

Further approximations are as in the first case

• We wish to [know] the measure of each side of the square that is five.

• We wish to know the measure of the sides of the square that is seven.

• We wish to know the measure of the square that is six.

Chapter Eleven – Word Problems

• Multiplication of a number by itself

• Multiplication of units and tens by units and tens
  

Word Problems

• Question: we summed all the successive numbers from one to twenty and it is the sum. How much is the sum?

• We want to know how much are the numbers summed up up to eleven.

There are other ways, but the above is the easiest and the right way

• Reverse question: the sum of the successive numbers starting from one is 210. What is the last number of the summed [numbers]?

איזהו המחובר האחרון מהמחוברים

If there is no number such the - the calculation is mistaken

• Question: a cane 5 cubits tall, is standing next to a wall of the same height.
  

If we lower its [top] two cubits down from the top of the wall, so that it will stand on a slope, how far will be the bottom end of the cane from the foot of the wall?

אם נשפיל אותה מראש הכותל אמתים כדי שתעמד בשיפוע כמה הרחיק ראש הקנה התחתון מיסוד הכותל

If the distance is a non-square number - using an approximation

• Question: I gave a messenger 30 dinar and one pašuṭ and ordered him to hire workers as much as his money allows, so that the payment of the one equals the payment of his friend, there will be no worker whose payment is one pašuṭ, and there will be no fractions in one's payment.
  

We want to know how many workers he could hire

נרצה לדעת כמה פועלי' יוכל לשכור

pešuṭim

19 workers for at 19 pešuṭim each

• Question: a wall collapsed. We rebuild it with an extension, so that it will be higher than what it was by half the size it had at the beginning, and its sixth, and ninth. With the whole extension, its height was 50 cubits. How much was its original height?
  

False Position:

Rule of Four:

cubits