Introduction: The Positional Decimal System

• I.e. the numeral that is in the first rank are the units.

• In the second - the tens.

• In the third - the hundreds.

• In the fourth - the thousands

Table of Contents

Section One: Integers

Chapter One: Addition

Written Addition

Description of the Procedure

• If you do not find there any number, but zeros, put a zero in the rank of the units.

• If you find a number or numbers with zeros, do not [pay attention to] the zeros and sum the numbers that are in that rank.

• If the [interim] result is ten or tens, without units, put a zero beneath the line, in the place of that rank, and keep the ten, or tens, as units to sum them with what you find in the succeeding rank.

In order that you will not forget them, put a dot or dots on top of the number in the succeeding rank, as the number of the reserved tens that are kept as units.

• If the [interim] result is ten or tens and units, put the number of the units beneath the line, in the place of that rank, and keep the ten, or tens, as units to sum them with what you find in the succeeding rank.

If you do not find in the succeeding rank but zeros, do not pay attention to them, since you have a [reserved] ten, or tens, to put as units in that rank. Put these tens as units beneath the line corresponding to that rank.

• If you do not have a ten, or tens, and you find in that rank only zeros, put beneath the line a zero corresponding to that rank, as mentioned for the first rank, when there is no number there but zeros, for there is one rule to this.

Example

• You wish to sum up two hundred and five thousand and five with three hundred ninety thousand and five and with six hundred twenty five thousand and two.

Set the digits as follows:

• First rank: Say: 3 and 5 are 8, plus 2 they are 10.

Since you do not have units, but a whole ten, put 0 beneath the line corresponding to the first rank.

Keep the ten, as one, for the succeeding rank, and put one dot above it so it will not be forgotten.

• Second rank: Since you do not find there any number, and you have that ten, do not pay attention to these zeros, put one under the line corresponding to the second rank, for the ten that you reserved as one.

• Third rank: Go to the third rank and since all of it are zeros and you do not have any reserved number, put one zero beneath the line corresponding to the third rank.

• Fourth rank: Go to the fourth rank and you will find there numbers and a zero. Since there is a number or numbers there, do not pay attention to the zero or the zeros that are there.

Say: 5 and 5 are 10.

Put 0 beneath the line, corresponding to that rank, as you did in the first rank.

Keep the 10 as one to sum it with what you find in the fifth succeeding rank and put one dot above it so it will not be forgotten.

• Fifth rank: Say: one for the reserved and 9 are 10, plus 2 are 12.

Put the two units beneath the line, keep one for the ten, and put one dot above the following rank.

• Sixth rank: Say: one for the reserved and 2 are 3, plus 3 are 6, plus 6 are 12.

Put the two units beneath the line, corresponding to that rank, and keep the ten as one in the succeeding rank.

Since the numbers are complete [and there is no more rank there] put that ten as 1 in the seventh rank, which is the succeeding rank.

Check

• Subtraction: subtract the first row from the result, subtract the second [row] from the remainder and so on, until only one [row] is left to subtract and the remainder then is equal to [the row] that you did not subtract.

As appears in the following diagram:

• Example:

For, when we subtract the row 625002 from their sum beneath the line, which is 1220010, 595008 remains, we subtract from this the other row, which is 390005, and 205003 remains, which is equal to the remaining row that was not subtracted yet.

Q.E.D.

Reason: Procedure

Reason: Check

Chapter Two: Subtraction

Written Subtraction

Description of the Procedure

• If you cannot subtract it from what is above it, as it is smaller than it or 0, take one from the upper digit that is in the succeeding rank, so it is ten in the [present] rank, and put one dot beneath the upper digit, from which you have borrowed the one.

• Even if there is 0 [in the succeeding upper rank], borrow from it and put a dot under it, so the one that you have borrowed is 10 in the [present] rank.

• Even if there is only 0 in this rank, subtract the bottom [digit] from it and put the remainder from these 10 under the line.

• If there is a number in the [present] rank, add the 10 to what you find there and subtract the corresponding bottom digit from the total sum, then put the remainder under the line, corresponding to the [present] rank.

• When you go to the rank, from which you have borrowed the one, where you have put the dot, add one for it to what you find in that rank in the bottom row, if there is a number there, and subtract the total from this rank in the upper row.

• If you do not find there enough to subtract, borrow one from the succeeding rank and put a dot under it, so it is ten in the [present] rank, as explained and so on.

• If there is no number on the bottom row in the place of the dot, for instance when there is a zero there, or nothing, as the bottom row is already complete, subtract the one from the rank, under which the dot is, if there is a number there, and put the remainder under the line, corresponding to the [present] rank.

• If there is no number but 0 in the [upper] row corresponding to the dot, borrow one from the succeeding rank and put under it a dot, so that this one is ten [in the present rank]. Subtract from it the one and put the remainder under the line, corresponding to the [present] rank.

• When the whole procedure is complete, if there are still digits on the upper row, under which there is no number, nor 0, or a dot, since all is complete, put them as a remainder under the line, as they are.

Example

• We wish to subtract 40438 the smaller from the greater number that is 76540304.

We put the smaller under the greater, like this:

This second diagram is as follows:

• First rank: We start to subtract from the first rank and say: 8 cannot be subtracted from 4. We borrow one from the succeeding upper rank, where there is 0. Put a dot beneath it. The one becomes a ten in the first rank and with the 4 that are in it they are 14. We subtract 8 from them. 6 remain. We put them under the line corresponding to the first rank.

• Second rank: We go to the second rank and we find there a dot. We add it to 3 that is found in this rank on the bottom row, they are 4. We cannot subtract them from 0 that is on the upper row in this rank, so we borrow one from the third rank and put a dot beneath it. So we have a ten, we subtract from them 4, 6 remain. We put them under the line, corresponding to the second rank.

• Third rank: We go to the third rank and add the dot to 4 that is on the bottom row, they are 5. We cannot subtract them from 3 that are above them, so we borrow one from the fourth rank, where there is 0, and we put a dot beneath it. This one becomes a ten, we add to it 3 that is on the upper row in the third rank and the total is 13. We subtract from them 5, 8 remain. We put them under the line, corresponding to the [third] rank.

• Fourth rank: We go to the fourth rank and we find there a dot, but there is no number in the fourth rank on the bottom row to add to it, only 0, so we subtract this one alone from what is found in the fourth rank on the upper row. Yet, we cannot since there is no number there but 0, so we borrow one from the fifth rank and put a dot beneath it. The 1 becomes a ten, we subtract from it 1, 9 remain. We put them under the line, corresponding to the [fourth] rank.

• Fifth rank: We go to the fifth rank and add the dot that is found there to 4 that is found in the fifth rank on the bottom row, they are 5. We cannot subtract them from 4 that above them, so we borrow 1 from the succeeding sixth rank and we put a dot beneath it. Thus, we have 10 and with 4 they are 14. We subtract from them 5, 9 remain. We put them under the line.

• Sixth rank: We go to the sixth rank and we find there a dot, but there is no number corresponding to it on the bottom row, since it is already complete, so we subtract the one alone from 5 that is in the sixth rank on the upper row and 4 remain. We put them under the line.

• The digits that are on the bottom row are already complete, as well as the dots, and there are two digits left on the upper row. We put them under the line as they are successively, as a remainder.

Thus, what resulted under the line is what remains from the greater number after we subtract the smaller from it.

Reason: Check

Check

Chapter Three: Multiplication

Written Multiplication

Description of the Procedure

• Multiply the first upper digit by each of the bottom digits and always write the units of the result of its product by each of the bottom digit beneath the line, under that bottom digit and keep the tens to add them as units to its product by the following digit to the left.

• If there is no number in the following rank of the bottom line since it is already completed or since there is a zero there, write the [reserved] number that you have in the following rank under the line.

• If when you multiply the upper digit by zero, you do not have [reserved] units, write a zero corresponding to that rank.

• If you have tens [reserved] as units, write them in that rank, under the line, as said, and do not write a zero at all.

• Once you complete multiplying the first top digit by each of the bottom digits, you start multiplying again the second top digit by each of the bottom digits and produce a second row from it.

Always write down the units and keep the tens as units to add them to the the next [product] as said.

• After you complete multiplying all the upper digits by the bottom [digits], draw an ink line beneath all these rows and sum them up, each with all its corresponding [digits], as we explained in the chapter on addition, and the result is the sought, which is the product of the two numbers one by the other.

• If there is a zero in the top row of the multiplicands, it would be expected to produce a row of zeros from it, but there is no need for that, since when you keep that every row starts from the rank that corresponds the rank of the upper digit by which you multiply all the bottom [digits] in that row and as you proceed, you always move one rank to the left - when you keep all that, you do not need to take care of the zeros [or to produce] rows of zeros from at all.

• If at the beginning of the top line you find zero or zeros before any digit, you have to write one zero for each zero of them below the line and complete [this row] by multiplying the next digit that follows them in the top line by all the bottom digits.

• If you want also, in order that you would not be mistaken [...] always this procedure itself, i.e. that you put one zero for every zero in the top row, even if they are in the middle, in the rank that corresponds to that zero and complete this row by multiplying [the upper next digit that follows them] by all the bottom digits.

Example

• Example: we wish to multiply the number 9007500 by another number 5400920.

Set the numbers one above the other successively, like this:

• Since there are two zeros at the beginning of the top row, write under the line two zeros corresponding to them as their number.

• Complete this row by multiplying the next upper digit that follows them, which is 5:

• Multiply the 5 by the zero that is in the first rank of the bottom number; it is zero. Place it after the two mentioned zeros, which is the third rank that [precedes] the fourth rank. It follows that its row starts from the third rank.

• Then say: multiply 5 by 2; it is 10. Place the zero after the mentioned zeros.

• Keep the 10 as one, to add it to the product of 5 by 9, which is 45; the total sum is 46. Place the 6 after all the zeros and keep the 4.

• When you multiply the 5 by the zero that follows the 9, do not put zero, since you have 4 in your hand, so place it instead after the 6.

• When you multiply the 5 by the following zero, [place] a zero after the 4, since you have nothing in your hand.

• Multiply 5 by 4; the result is 20. Put a zero and keep the 2.

• Multiply 5 by 5; they are 25 and with the 2 that you have the total sum is 27. Put 7 and after it put 2, because the multiplication of the upper digit by all the bottom digits is complete.

• Since you find [a zero] after the upper 5 that was multiplied, the rule requires to multiply it by all the bottom digits and make of it one line.

But, there is no need for this, only that you start with the 7 that is after it and make its line provided that the line of the 7 starts from the fifth rank, which is its rank as mentioned.

In order not to be mistaken, place a zero at the beginning of the line of the 7, for the upper zero that is between the 5 and 7, whose rank is the fifth rank.

Complete the line by multiplying the 7 that follows it by all the bottom digits:

• Say: the product of 7 by zero is zero. Put another zero.

• Say also: the product of 7 by 2 is 14. Place 4, which are the units, after the mentioned zeros and keep the 1.

• Say also: the product of 7 by 9 is 63; plus the one that we have the result is 64. We put the 4 and keep the 6 as units.

• We say also: the product of 7 by zero is zero. Since we have a 6, we do not put a zero, but we put the 6 that we have in its place.

• We say: the product of 7 by zero is zero. We put a zero, as we do not have units at all.

• We say: the product of 7 by 4 is 28. We put 8 and keep the 2.

• We say: the product of 7 by 5 is 35; plus the two that we have the result is 37. We put the 7 and after it the 3, because the multiplication of the 7 by all the bottom digits is complete.

• Since there are two zeros after the 7 in the upper line, in order not to be mistaken, we put two zeros corresponding their rank at the beginning of the line of the 9 that follows them.

• Again we say: the product of 9 by zero is zero. Put a zero after the two zeros that we placed at the beginning of this line.

• We say: the product of 9 by 2 is 18. We put 8 after the three mentioned zeros and keep the one.

• We say: the product of 9 by 9 is 81 and since we have 1, the result is 82. We put the 2 and keep the 8.

• We say: the product of 9 by zero is zero. We put the 8 that we have in its place.

• We say: the product of 9 by zero is zero. We put it, as we have nothing in our hands.

• We say: the product of 9 by 4 is 36. We put 6 and keep the 3.

• We say: the product of 9 by 5 is 45; plus the 3 that we have the result is 48. We put the 8 and after it the 4, because the multiplication of the 9 by all the bottom digits is complete.

• Since the upper digits were already multiplied by all the bottom [digits], we draw a line beneath all the lines and sum all the lines that were generated from their products, i.e. the three lines between the lines.

We receive that the result of the multiplication of the two questioned numbers one by the other is 486463564860000.

Check

• Division: If you want to check if you did it right or not, divide the great number resulting from the multiplication by one of the two multiplied numbers, so the result of division is the other [multiplied number]. But if it is lacking or exceeding, know that you have erred in one of the operations of multiplication or division.

Reason: Procedure

• Because, e.g. if the upper number is of the hundreds, which is the third rank, when we multiply it by the units of the bottom number, the result is of the hundreds.

• If it is of the thousands, which is the fourth [rank], the result is of the thousands.

Therefore, the units resulting from the first multiplication are in the third rank, which corresponds to the rank [of the upper number], and the tens resulting from this multiplication are units of the subsequent rank, as explained in the beginning of the book, in the explanation of the ranks. So, you keep them as units to add them to the next [product].

• When we multiply the upper number by the second of the bottom [digits], the result are tens of hundreds, which are thousands, if the upper [number] is of the hundreds.

• If it is of the thousands, these tens are tens of thousands, so we place them in the rank that follows the one that we place at the beginning of this line and we add to them the reserved from the previous multiplication.

And so on, forever it rises rank by rank, until it is clear from what is said that the place of the units resulting from the multiplication of each upper digit by a bottom [digit] is the rank whose decimal position is as the [sum] of the ranks of both upper and bottom digits that are multiplied by each other minus one.

• If one of them is in the first rank, we have explained that the place [of the product] is in the rank that corresponds the rank [of the other multiplied digit]

• If it is in the second rank, [the product] rises one rank above the rank of the other [multiplied] digit as explained.

• If it is in the third [rank], [the product] rises by two.

And so on it exceeds over the ranks of the second [multiplied] digit by the number of the ranks [of the first multiplied digit] minus one.

Hence, the rank of the units resulting from the multiplication is as [the sum of] the ranks of both digits that are multiplied one by the other minus 1 and all this is clear by reason.

Reason: Check

• The total [amount of] money is the product of the number of people multiplied by the amount of money that each of them receives.

• Or, if you want to say: the product of the money that each of them gets multiplied by the number of people. All is the same.

For, twelve is a product of 3 by 4, or 4 by 3.

• Thus, if we divide these 12 between 4 people, each receives 3

• and if between 3 [people], each receives 4

Chapter Four: Division

written division

description of the procedure

• See how many times the last bottom digit to the left can be extracted from the last digit of the upper [number].

• If it is not even once in it, as it is smaller than it, see how many times it can be extracted from the last [digit] and the one that precedes it, considering the last [digit] as tens and the one that precedes it as units. The number of these times is called the result of division.

Know that you have to extract the digits that precede the last bottom digit as many times from the upper one that precedes the last digit or digits, from which you extract the multiples of the last bottom digit, as the number of times that you extract the last bottom digit from the last upper digit or digits.

• When there is a remainder from the last digit or digits, consider them as tens or hundreds, according to their relation to the present rank, [subtract from them] as the multiples that you subtract the last bottom digit from the last upper digit or digits.

• If there is not enough, do not subtract from the last digit as much as these multiples, for you always have to subtract each [bottom digit] from its corresponding upper digit as many times as you subtract the last [bottom digit] from the last [upper] digit or digits.

• You should know that whenever you wish to subtract the bottom digit from the upper digit and you do not find in its corresponding rank enough [to subtract] its multiples, if there is in the one that follows the consecutive [rank], you can subtract from it, provided that you keep the positional value of the ranks, so that every digit is tens to its preceding and hundreds to the one that precedes its preceding and so on according to this relation.

• Interim result: After you know the multiples of which you can subtract each of the bottom digits from the corresponding upper rank or ranks,

i.e. for example, if the last bottom [digit] is subtracted from the [seventh] [upper] rank and the one that precedes the last is subtracted from the sixth [rank] and the one that precedes the preceding is subtracted from the fifth, until they are complete.

in the rank that they end, i.e. where the first bottom digit is to be subtracted as many times from that rank according to the order explained - place the number of the multiples that are subtracted corresponding to that rank, beneath the upper digit.

• Interim remainder: When there is a remainder from the upper digit, place the remainder above it, so that this remainder will be in front of your eyes to be used as tens or hundreds for the preceding [digit] and the one that precedes the preceding, as explained.

• Repeated division: When all the bottom [digits] to be subtracted from their corresponding [upper digits] are complete, if there is a remainder in the upper number that is as the bottom number or more, we divide it again as in the beginning: We see how many times the last bottom digit can be subtracted from the last digit or digits of this remainder, as we have done in the beginning with the whole number, and the preceding and the one that precedes the preceding, all by the same multiples, each from its corresponding [upper digit].

• Last interim division: We always repeate [the division] time and time again, until the time comes when each of the bottom [digits] are subtracted from its very rank, i.e. the units from the units, the tens from the tens and the number of multiples is placed at that time in the first rank.

We do not divide again, because what remains then is less than the bottom number.

The less cannot be divided by the more into integers but only into fractions.

We will mention further in this chapter how it will be divided into fractions.

• The decimal place of the interim result: Always remember to place each time the result of division of that time, i.e. the multiples that you subtract at that time, corresponding to the rank from which the first bottom digit is subtracted, i.e. the [digit] that is in the first rank, if there is a number there, or if there is only a zero there - see from where it should be subtracted, if there is a number there, and place there the result of division of that time.

It turns out that when we want to know from which rank a certain bottom digit should be subtracted, we should see from which rank the last bottom digit should be subtracted at that time, then count the number of the ranks from there to the right, as the number of the ranks that this bottom digit is far to the right from the last bottom digit - where they end it should be subtracted from the upper digit.

Likewise, when you wish to know in which rank to place the result of division each time, see from which rank the last bottom digit was subtracted at that time, count from there to the right, as the number of the bottom digits, and where this counting ends, place the result of division at that time. From that very rank the digit that is in the first rank of the bottom number is subtracted at that time.

example

We put them in two spaced lines orderly one on top of the other like this:

• We say: 4, which is the last bottom digit, can be subtracted once from the last upper digit.

But, we cannot subtract 6 that precedes the last bottom digit from 4 that precedes the last upper digit

Therefore, we do not subtract from there [= the last upper digit], but we subtract from the last two [upper] digits, which are 43.

We say: how many times 4 is in 43?

We do not say ten, since if the 3 that in this rank could have taken 10, it would have taken it from the last digit, which is ten in relation to this rank.

Therefore, we say: only 9.

The bottom line has 5 digits, hence we count 5 ranks to the right from the upper 3, from which we subtract. They end in the zero and we place beneath it the 9 resulting from the division, which is the number of times that we have to subtract the bottom digits from the upper digits at this time, each from its corresponding rank, as mentioned.

• We say: 9 times 4 are 36.

We have to subtract the 6 units from the 3 and the 3 tens from the 4.

There is not enough in 3 to subtract the 6 units from it, so we take a ten from the 4 and add it to the 3 tens, from which we should subtract.

We take the whole 4 and place above it a zero, or cross it out by a pen.

We add the additional ten that we took to the upper 3; they are 13. We subtract from them the 6 units that we have to take from there; 7 remain. We place them above the 3.

Thus, we already took the units and the tens, each from its corresponding rank.

• We say also: 9 times 6 are 54.

We subtract the 4 units from the rank that precedes the last mentioned upper [digits], which is the 8; 4 remain. We place them above it.

We subtract also the 5 tens from the rank that follows it on the left, which are tens in relation to it; you find there 7. Subtract 5 from it; 2 remain. We place them above it.

• Also 9 times 7, because we do not pay attention to the bottom zero, except for keeping the ranks, i.e. it indicates for us that the multiple of 7 that precedes it should not be subtracted from the rank of the upper zero that precedes the 8, from which we subtract the 6, but from the rank of the 4 that precedes the 8 by 2 ranks, as the 7 precedes the 6 by 2 ranks.

We say: 9 times 7 are 63.

We subtract the 3 units from the upper 4 mentioned; 1 remains. We place it above it.

We should have taken the 6 tens from the rank of the zero, but as there is not enough there to subtract them, we take one from the following rank. You find there 4. We take one; 3 remain. We place them above it.

This one is ten in relation to the rank of the upper zero, from which we take the tens.

Since each of these ten is ten in relation to the 4, from which the 7 is taken.

We subtract the 6 tens from these 10; 4 remain. We place them above the zero.

• We say also: 9 times 9 are 81.

We should have taken the the one from the rank of the upper zero that precedes the 4, from which we took for the 7, but since there is not enough there, we take the one that is above the 4, so it is ten here. We subtract the one; 9 remain. We place them above it.

We should subtract the 8 tens from the rank of the 4, but there is nothing there, since even the one that was there is already taken. Therefore, we take one from the 4 that above the zero that follows; 3 remain. We place them above it.

The one becomes a ten above the 4, from which we subtract the 8 tens and each unit of these tens is ten in relation to the rank of the zero, from which the [1] is subtracted. We subtract the 8 tens from these ten tens; 2 remain. We place them above the 4, i.e. above the one that is above the 4.

The same multiples of all the bottom digits were subtracted, each from its corresponding [rank], as the last bottom digit was subtracted from the last [upper] digits.

We are left with 233298998 in the upper number, which is greater than the bottom number by some multiples.

Therefore we continue to divide this remainder by it as in the beginning and write a line above the remainder, so we will not be confused.

• We say: the last bottom 4 cannot be subtracted even once from the last remaining two, therefore, we subtract it from the last 23. It is there five times.

We place this 5 beneath the first that is 5 ranks to the right from the 3, from which we subtract, which is as the number of the ranks of the bottom line.

• We say: 5 times 4 are 20.

Since there are no units there, we do not subtract a thing from the 3, but we subtract these 2 tens from the 2 that follows. Nothing remains.

• We say also: 5 times 7 are 35.

We subtract the 5 units from the 9 that is above the 0, since this is the place from where they are subtracted, in the rank of 4, to the right of the 3 that is above the upper 8, from which the last bottom digit was subtracted, as the distance of the 7 to the right from the last bottom 4. After we subtract the 5 units from the 9 mentioned, 4 remain. We place them above it.

We cannot subtract the 3 tens from the following rank, therefore we take one from the 3 that follows; 2 remain. We place them above it.

This 1 is ten with relation to the two. We add them to is; they are 12. We subtract 3 tens from them; 9 remain. We place them above it.

• We say also: 5 times 9 are 45.

We subtract the 5 from the 8; 3 remains.

We subtract the 4 tens from the 4; nothing remains. We place 0 above it.

All the bottom digits were subtracted and we are left with 2943998 in the upper number, which is much greater than the bottom number.

Therefore we continue to divide them and draw a line above the remainder, so we will not be confused.

• We say: 4 cannot be subtracted from 2, yet they can be subtracted from 29. They are subtracted from it 7 times and only 1 remains, but it is not enough for the tens of the product of 7 by 6 that precedes the 4, which are 4 tens.

Therefore, we take only 6 and place it beneath the 9, which is 5 [ranks] to the right of the 9, corresponding to the upper [digit], from which we subtract the last bottom digit, i.e. as the number of the ranks of the bottom line.

Since there is a rank that is empty from a number between the 5 that resulted from the division and the current result, we place 0 in the empty rank, for this is the role of the 0, as we explained at the beginning of the book.

• We say: 6 times 4 are 24.

We subtract 4 units from the 9; 5 remain. We place them above them.

We subtract 2 tens from the 2 that follows; nothing remains. We cross it out with a pen.

• We say: 6 times 6 are 36.

We should have taken the 6 units from the 0 that is in the remainder, but we cannot, therefore we take one from the 5, from which we subtract the 3 tens.

It is ten [in the present rank]. We subtract 6 units; 4 remain. We place them above it.

We subtract also the 3 tens from the 5, but we already took 1, so only 1 remains there. We place it above it.

• We say also: 6 times 7 are 42.

We subtract the two units from the 9, i.e. in the rank of hundreds, which is 4 ranks to the right of the rank from where the last bottom digit was subtracted; 7 remain. We place them above it.

We should have taken the 4 tens from the 3, but is not enough, therefore we take one from the 4 that follows; 3 remain. We place them above it.

This 1 in 10 in the rank of the 3. We sum them; they are 13. We subtract the 4 tens from them; 9 remain. We place them above them.

• We say: 6 times 9 are 54.

We subtract the 4 from the 9; 5 remain. We place them above them.

All were subtracted and 139258 still remains in the upper [number], which is more than the bottom [number].

Therefore we continue to divide them and draw a line above the mentioned remainder.

• We say: in 1 there is not enough for 4, but we subtract from the last 13 of the remainder. It is subtracted 3 times. We place the 3 that is found in division beneath the upper 8, which is in the first rank, 5 [ranks] to the right of this 3 from which we subtract, as the number of the ranks of the bottom line.

• We say: 3 times 4 are 12.

We subtract the 2 from the 3; 1 remains. We place it above them.

We subtract 1 from the 1; nothing remains. We cross it out with a pen.

• We say: 3 times 6 are 18.

We subtract the 8 from the 9; 1 remains. We place it above them.

We subtract 1 from the 1; nothing remains. We cross it out with a pen.

• We say: [3] times [7] are [21].

We subtract the 1 from the 5; 4 remain. We place them above them.

We subtract 2 from the 2; nothing remains. We cross it out with a pen.

• We say: 3 times 9 are 27.

We subtract the 7 from the 8; 1 remains. We place it above them.

We subtract the 2 tens from the 4; 2 remain. We place them above them.

All were subtracted and 1021 remain, which is less than the bottom [number], therefore cannot be divided by it into integers.

So you may say that you have finished the procedure completely.

• The result of division is 95063 for each.

• The remainder in the upper number that cannot be divided into integers is 1021.

check

Multiply the result of division, i.e. 95063, by the divisor, i.e. 46079, add the remainder, i.e. 1021, to the product and you will receive the dividend, i.e. 4380408998, which is the number that you divided.

If the result is not as the [dividend], know that you made a mistake in one of the operation, i.e. the division or the multiplication.

You will find it alluded in the noted diagram.

• In order that you will learn it, I write here the check of the example that I have brought in the third chapter of multiplication, that is we divide the result of that multiplication, which is 486463564860000 by one of the two multiplied numbers, first it will be by 5400920.

We arrange them one on top of the other, the greater above, like this:

• We say: 5 cannot be subtracted from 4, but it can be subtracted from 48. It is found there 9 times. We place the 9 beneath the 6, which is seven ranks to the right of the 8, as the number of the ranks of the bottom line.

• We say: 9 by 5 are 45.

We subtract the 5 from the 8; 3 remains.

We subtract the 4 from the 4; nothing remains.

• We say also: 9 by 4 are 36.

We subtract the 6 from the 6; nothing remains.

We subtract the 3 from the 3; nothing remains also.

• We say also: 9 by 9 are 81.

To know from which rank we subtract, do one of the two things:

• Count from the rank where you put the result of division, which is the 6, three ranks to the left, as the distance of the 9 from the first rank. They end in the upper 3 and from it we subtract the one resulted in the multiplication.

• If you want, count from the rank from which the last bottom digit was subtracted, which is the upper 8, five ranks to the right, as the distance of the 9 from the last digit 5. They end also in 3.

We subtract the 1 from there; 2 remains.

• We say also: 9 by 2 are 18.

• We say: 7 by 5 are 35.

We subtract the 5 from the 8; 3 remains.

The 3 from the 3; nothing remains.

• We say: 7 by 4 are 28.

• We say also: 7 by 9 are 63.

• We say also: 7 by 2 are 14.

We subtract the 4 from the 8; 4 remains.

We subtract the 1 from the 1; nothing remains.

• We say: 5 by 5 are 25.

We subtract the 5 from the 7; 2 remains.

We subtract the 2 from the 2; nothing remains.

• We say: 5 by 4 are 20.

• We say also: 5 by 9 are 45.

reason: procedure

• I.e. if we divide six hundred by 3 units:

Each unit receives 2 and the rank from which it is subtracted is the rank of hundreds, so the two resulting from the division are of the same rank and they are two hundred.

If there were thousands in the upper [line] and tens in the bottom [line], then as the units receive from the hundreds, so the tens would receive from the thousands.

Because, as the tens are tens in relation to the units, so the thousands are tens in relation to the hundreds; and as the hundreds are in relation to the units, so the thousands are in relation to the tens, and the tens of thousands are in relation to the hundreds; and so on, as those are in relation to those, so are these in relation to these.

We have already explained that the rank of the [number of] times that the units [are subtracted] is the rank from which they are subtracted

[Thus, since the units are subtracted from the hundreds], the tens are subtracted from the thousands, and the hundreds from the tens of thousands and so on.

The rank of the hundreds is third to the left of the units, so when the units are taken from the hundreds, as we mentioned, the tens should be taken from the thousands that are third [in relation to the tens] to the left of the rank of hundreds, from which the units are taken.

reason: check

• If we divide 20 by 5, each one receives 4; if by 4, each one [receives] 5.

So, 20 is a product of 4 by 5, which are the number by which we divide and the result of division.

• If we divide 21 by 4, each one receives 5 and 1 remains.

Therefore, when we multiply 4 by 5, the result is 20. We add to it the remaining 1; the total result is 21, which is the same as the dividend.

divisibility of a number

3; 6; 9

• six: See, if the first digit that is in the first rank is an odd number, then you know that the number does not have a sixth [= not divisible by 6].

If it is an even number, know that if it has a third [= divisible by 3], it has also a sixth [= divisible by 6], otherwise it has not.

• nine and three: Every [number] that it has a ninth [= divisible by 9], has also a third [= divisible by 3], but not vice versa.

To know if a number has a ninth or a third, consider all the digits of the number as if they are of the first rank, i.e. sum them as units and cast out the nines from this sum

• No remainder - If it is consumed by the nines, know that it has a ninth [= divisible by 9] and it certainly has a third [= divisible by 3].

• The reminder is 6 or 3 - if 6 or 3 remains, it has a third [= divisible by 3], but it does not have a ninth [not divisible by 9].

• The reminder is a number other than 3 - if another number remains, such as 4 or 5 and the like, it does not even have third [= not divisible by 3].

• The reason that we consider all the digits of the number as units, without considering their ranks, is that every [unit of] a certain rank is ten [units of] the preceding rank, therefore when subtracting nine from the ten [of a certain rank] the remainder belongs to the preceding rank, and so on repeatedly. We find that after casting out the nines [their ranks] are the same.

נמצא שלאחר הסרת התשיעיות כלם שוים

• We say that if the first digit is an odd number, it does not have a sixth [= 6 is not a divisor of the number], since the whole number is odd, and an odd number is indivisible by an even number, i.e. it is not a product of an even number multiplied by an odd number and all the more so by an even number.

• We say that if it is an even number that has a third [= divisible by three], it has also a sixth [= divisible by six], since the whole number that consists of an even number multiplied by 3 is an even number. For if it were an odd number, it were a product of an odd number by an odd number, but as the number that consists of the 3 is even, it is divisible by double the 3, i.e. by six, so it has a sixth.

2; 4; 8

• two: if it is an even number or 0 - then the whole number is even, for the tens and up are even - therefore it is known that it has a half [= divisible by 2].

• four and eight: to know if it has also a quarter and an eighth:

The digits of the whole number are summed according to the following procedure:

• Take the numeral that is in the first rank as it is.

• Double what is in the second [rank], if there is a number there.

• If what is in the third [rank] is an odd number, take 4 units for it; if it is an even number or 0, do not take any thing for it.

• Likewise from the third rank up, do not take any thing.

Sum up all that you took and cast out the eights.

• If all is gone [= no remainder], it has an eighth and a quarter [= divisible by 8 and 4].

• If four remains, it has a quarter [= divisible by 4].

• If another number remains, it does not have an eighth nor a quarter [=not divisible by 4 or 8]

7

• If there is no remainder - the number has a seventh [= divisible by 7]

• Otherwise - it does not

5; 10

• If the digit of the units is zero - the whole number is tens and therefore it has a tenth and a fifth [= divisible by 10 and 5]

• If the digit of the units is 5 - the number has a fifth [= divisible by 5]

• If the digit of the units is other than 0 or 5 - it does not have a fifth [= not divisible by 5]

11

• If there is no remainder - the number has an 11th [= divisible by 11]

• Otherwise - it does not

13

• If there is no remainder - the number has a 13th [= divisible by 13]

• Otherwise - it does not

general rules

• If the numbers 2-11, 13 are not [divisors] of a given number - any [divisor] of the given number is not divisible by these numbers

The reason: if one of these numbers was a [divisor] of the [divisor], then this [divisor] could not have been a [divisor] of the given number, for if it were a [divisor] of the given number then its own [divisor] should have been the [divisor] of the given number also

• 21 is a [divisor] → 7 and 3 are also [divisors]

• If 1; 2; 3; …; n-1 are not [divisors] of a given number a, and a<n² - then n is not a [divisor] of a

The reason: if n is a [divisor] of a and a<n², then there is b<n, so that a=b·n, but b cannot be a [divisor] of a according to the condition that 1; 2; 3; …; n-1 are not [divisors] of a → contradiction

• 1; 2; 3; …; 16 are not [divisors] of a → if a is divided by 17, 17 times or more, then a≥17² = 289; and if it is divided by 17, 16 times or less, then one of the numbers smaller than 17 is a [divisor] of a → contradiction

ואם הפעמים האלה אשר י"ז בו הם פחות מי"ז המשל י"ו ומשם ולמטה הנה היה לו רביעית או אחד מהמורים הקודמים שהרי יתחלק למספר פעמים אלו ג"כ ויצא בחילוף הי"ז וזה שקר ‫^[81]שהרי לא מצאת לו אחד מהעוברים

• If 1; 2; 3; …; n-1 are not [divisors] of a given number a, and a≥n²

• If there is an integer m, so that a÷n=m → then n and m are [divisors] of a

• Otherwise - n is not a [divisor] of a

Repetitive division of a number by its [divisors]

• Example: we wish to divide 2447235 by 50335084800.

We extract the divisors of this large number, by which we want to divide.

• First, we see if 3, or 6, or 9 are [its divisors]:

Since the first digit is 0, we know that if 3 or 9 [are its divisors], then 6 is also its [divisor].

To know if 3 or 9 [are its divisors], we sum up all [the digits] as if they were units.

We say: 8 with 4 is 12. We subtract from it 9; 3 remains.

With 8 it is 11. We subtract the 9; 2 remains.

With 5 it is 7 and with 3 it is ten. We subtract 9; 1 remains.

With 3 it is 4 and with 5 it is 9. So, the number is cast out by nines.

Hence, we know that 9, 3 and 6 are its [divisors] for the reason we have mentioned.

We take whichever of them we want.

• For instance, we take 6 as its divisor. We divide it by 6; the result of division is 8389180800.

Since the result of division is a large number, we seek its divisor. We do as we did with the original number; it is all cast out by nines.

Since the first digit is a zero and 9 is its divisor, 6 is also its divisor and all the more so is 3.

• We take 9 as its divisor and divide this number by 9; the result of divison is 932131200.

We do with it as we have done with the formers; 3 remains.

So, 3 is its [divisor] and 6 also, since the first [digit] is a zero.

• We take 3 as its divisor and divide it by it; the result of divison is 310710400.

Since 9 is not a [divisor] of the former, it is known that even 3 is not [divisor] of this [number].

The reason is that this [number] is a third of the former [number], so if this [number] had a third, its third, i.e. the third of the [number] that is a third of the former [number], is a third of a third of the former [number], which is a ninth, but we have found that it does not [have a ninth].

So it is truly, when you sum up [the digits], then cast out the nines, 7 remains.

We see if 2, 4, or 8 are its [divisors].

Since the first [digit] is a zero, it is known that 2 is its [divisor].

To know if 4, or 8 are its [divisors], we should take what is in the first rank. We find there only 0, so we take nothing. We take what is in the second [rank]. We find there only 0, so we take nothing. Since the third [rank] is an even [number], we do not have to take anything for it or for the higher [ranks]. So, 8, 4, and 2 are its [divisors].

הנה יש לו ח' וד' וב‫'

We take whichever of them we want as a divisor.

• For instance, 8. We divide it by 8; the result of division is 38838800.

It also has an eighth, a quarter and a half, for the reason mentioned.

We take one of them.

• For instance, 4. We divide by it; the result of division is 9709700.

It does not have an eighth, because the first [rank] and the second [rank] are 0, so we take nothing for them; but the third [rank] is 7, which is an odd number, so we take 4 for it. Hence, only 4 and 2 are its [divisors].

הנה שאין לו כי אם ד' וב‫'

We take one of them.

• For instance, 2. We divide it by it; the result of division is 4854850.

Since this [number] is half the former [number], it does not have a quarter, because the former [number] does not have an eighth, as a quarter of this [number] is a quarter of half the former, which is an eighth.

You will also find by examination that it does not have a quarter: since there is no number in the first [rank], the second is ten, and the third is an even number, so we take nothing for it. Therefore, we have only ten. We subtract 8; 2 remains. Hence, it has only a half.

If you do not want to take it again as a divisor, see if 5 or 10 are its [divisors].

Since the first [digit] is 0, it is known that 10 as well as 5 are its [divisors].

We take one of them.

• For instance, 5. We divide it by it; the result of division is 970970.

Since the first [digit] is 0, 10 and 5 are its [divisors] also.

We take the ten, for instance, and divide it by 10; the result of division is 97097.

Since the first [digit] is not 5 nor 0, neither 5, nor 10 are its [divisors].

We see if it has a seventh.

We subtract seven from the last 9; two remains.

We multiply it by 3; it is 6. We add it to the 7 that precedes it; it is 13. We subtract 7; 6 remains.

We multiply it by 3; it is 18. We subtract 14, which are two sevens, from it; 4 remains.

Since there is no number in the preceding rank, only 0, we multiply the 4 by 3; it is 12. We subtract 7; five remains.

We add it to the preceding [digit]; it is 14, which are sevens. The first [digit] is also 7; so all is cast out by sevens.

It has a seventh.

• We take it as a divisor and divide it by it; the result of divison is 13871.

We see if 7 is its [divisor] also.

We find that 4 remains, so 7 is not its [divisor].

We see if 11 is its [divisor].

We subtract the last 1 from the 3 that precedes it; 2 remains.

We subtract it from the 8 that precedes it; 6 remains.

We subtract it from the 7 that precedes it; 1 remains.

We subtract it from the 1 that precedes it; nothing remains.

11 is its [divisor].

• We take it as a divisor and divide it by it; the result of divison is 1261.

We see if 11 is its [divisor].

We find that 7 remains, so 11 is not its [divisor].

We see if 13 is its [divisor].

We multiply the last 1 by 3, but we cannot subtract it from the 2 that precedes it, so we add 13 to it; it is 15. We subtract the 3; 12 remains.

We multiply it by 3; it is 36. We subtract 26, which are 13 and 13; 10 remains.

We cannot subtract it from the 6 that precedes it, so we add 13 to it; it is 19. We subtract the 10 from it; 9 remains.

We multiply it by 3; it is 27. We subtract 26; 1 remains.

We subtract it from the first 1; nothing remains.

Hence, 13 is its [divisor].

• We take it as a divisor and divide it by it; the result of divison is 97.

Since this number is smaller that the square of 13, you do not need to see if 13 is its [divisor], all the more so a greater divisor, or one of the preceding [divisors], as they are not [divisors] of the former [number], so we take it as a divisor itself.

We have received the divisors of this number.

The smaller number that we want to divide by this greater number is 2447235.

We receive that when we divide 2447235 by [50335084800], the result of division is two-fifths of a half of a quarter of an eighth of a third of a ninth of a sixth, i.e. when we decompose one unit to 6 parts, then 1 of these 6 [parts] to 9, then 1 of these 9 [parts] to 3, then 1 of these 3 [parts] to 8, then 1 of these 8 [parts] to 4, then 1 of these 4 [parts] to 2, then 1 of these 2 [parts] to five, each of the units of the large number, by which we divide, receives two parts of the former parts.

We also get from the division 5-tenths of these parts, i.e. of a fifth of a half of a quarter of an eighth of a third of a ninth of a sixth.

Also one-seventh of a tenth of a fifth of a half of a quarter etc.

Also 4 parts of 11 of a seventh of a tenth of a fifth etc.

Also 9 parts of 13 of 11 [parts] of a seventh of a tenth of a fifth etc.

Also 22 parts of 97 of 13 [parts] of 11 [parts] of a seventh of a tenth of a fifth etc.

Example in our number: the dividend has a ninth. Write this divisor last to the left and divide by it first; nothing remains to write beneath it.

We see if the result of division has also one of the remaining divisors. Write it before the one by which we have divided and divide by it; nothing remains.

Thus we always do in this matter, so that we get, when we divide the mentioned small number by the greater number, the result that each unit receives a half part of 13 of [a part of] 11 of a quarter of a third of a sixth, plus 9 parts of 97 of a tenth of an eighth of a half etc. as it is in this diagram:

The dividend has a ninth, therefore we place the 9 last and divide by it; nothing remains, so we do not write any thing beneath it. The result of division is 271915.

This result of division has a fifth, therefore we place the 5 before the 9 and divide by it; nothing remains. The result of division is 54383.

It has a seventh, therefore we place the 7 immediately before the mentioned divisors and divide by it; nothing remains. The result of division is 7769.

It has none of the remaining divisors, therefore we place whichever we want. We place the greater, which is 97, and divide by it; 9 remains. We place it beneath and the result of division is 80.

It has a tenth, therefore we place the 10 immediately and divide by it; nothing remains. The result of division is 8.

It has an eighth, therefore we place the 8 immediately and divide by it; nothing remains.

Since it is smaller than any of the remaining divisors, we should not divise further, but we write it beneath the divisor that we write immediately after the mentioned written [divisors]: We write 2 and beneath it the 1 that results lastly in the division. Then: we write the remaining divisors randomly.

• We wish to divide 3123740520 by 216.

The divisors of the smaller number, by which we want to divide, are 9, 8, 3.

We divide our greater number by 9; nothing remains and the result of division is 347082280.

We divide it by 8; nothing remains and the result of division is 43385285.

We divide it by 3; the result of division of the last [number] is 14461761, which are the integers and 2 remain.

We write it beneath the 3 and these are the fractions resulting from the division that are added to the integers.

If now in this last division nothing would have remained also, we would not have written anything beneath it, and since nothing would have been found beneath the divisors, no fractions would have been resulted in division at all, only integers.

This happen when the number by which we divide is a divisor of the original number.

But, in our example, since 2 remains in the last division, it have also fractions.

We write it beneath the 3 and we get that when we divide 3123740520 by 216, each unit receives 14461761 integers and 2-thirds, according to the following diagram:

In the mentioned example, if we want to know if the resulting divisors are correct: we multiply 9 by 8; it is 72. We multiply it by 3; the product is 216. So, the divisors are all gone and the result is the number itself that have these divisors.

• Example in the following diagram:

We multiply the integers by 3, which is the first divisor; the result is 43385283.

We add to it the 2 that is beneath this divisor; the total is 43385285.

We multiply it by 8, which is the second divisor; the result is 347082280.

Since we do not find anything beneath this divisor, we do not add anything to it.

We multiply it again by 9, which is the third divisor; we receive the dividend itself, which is 3123740520.

• I shall present it briefly in the example that precedes this one, in which there are no integers in the result of division, only fractions:

We take the one that is beneath the 2, which is the first divisor beneath which there is something, and we multiply it by 8, which is the next divisor; the result is 8.

Since there is nothing beneath it, we do not add anything to it and multiply it by 10, which is the next divisor; the result is 80.

Since nothing is found beneath it, we multiply it without adding by 97, which is the next divisor; the result is 7760.

We add to it the 9 that is beneath it; the result is 77[69].

We multiply it by 7; the result is 54383.

We multiply it by 5; the result is 271915.

We multiply it by 9, which is the last divisor; we receive the dividend itself, which is 2447235, so it is correct.

• I.e. if 1 is divided by one hundred, for instance, each 1 receives from the hundred one part of the hundred that is in it.

Meaning, we decompose the one unit that we want to divide to one hundred equal parts and each 1 of the hundred, by which we want to divide the original 1, receives one part of them.

Since the 100, by which we want to divide, has a fifth and its fifth is 20, we find that each 1 of the hundred is one part of 20 of a fifth of a hundred.

When we decompose the whole unit to 100 parts, each one of them is one part of 100 of the whole and it is also one part of twenty of a fifth of the whole. So our saying one part of 100 of the whole is as our saying one part of 20 of a fifth of the whole.

For this reason itself, our saying a quarter of a fifth is as our saying one part of 20. Since the twenty has a fifth and its fifth is 4. We find that twenty consists of 5 and 4 and one [part of twenty] is a fifth of a quarter, or a quarter of a fifth, because it is the same. Hence, our saying a quarter of a fifth is as our saying one part of 20 of the whole.

But, our saying one part of 20 of a fifth of the whole is as our saying one part of 100 of the whole, as we have explained. So, we get that our saying a quarter of a fifth of a fifth is as our saying one part of 100 of the whole.

Thus, if we divide the whole unit by 100, each 1 of them receives one part of 100 of the whole, which is a quarter of a fifth of a fifth.

• If we divide 2 integers by 100.

Each one receives 2-quarters of a fifth of a fifth.

• If 3, 3.

• Thus, as the number of integers divided by 100, so is the number of quarters of a fifth of a fifth that each of them receives, and this is clear.

• If we wish to divide 70 by 100.

We know that each of them receives 70 quarters of a fifth of a fifth. It is as if we write the denominators by this order and place the 70 beneath the last denominator.

Since you have 70 quarters of a fifth of a fifth, if we want to know how many fifths of a fifth they are, it is as if we have 70 quarters and we want to know how many integers they are.

This is known by dividing them into 4, because every 4 quarters are one integer and also every 4 quarters of a fifth of a fifth are one-fifth of a fifth. So, we divide these 70 quarters of a fifth of a fifth and the remainder is of the first type, i.e. quarters of a fifth of a fifth.

Therefore, we should write the result of division, which is 17, beneath the 5, which is the preceding divisor and the remainder, which is 2, beneath the 4, which is the last divisor by which we divide, like this:

We know that when we divide 70 by 100, each one of them receives 17 fifths of a fifth and 2-quarters of a fifth of a fifth.

Since we have 17 fifths of a fifth, we know that they are three-fifths and more, because every five fifths of a fifth are one-fifth, as five-fifths are one integer.

We should know how many fifths they are, so we divide them by 5; the result of division is 3, which is 3-fifths, so we write it beneath the first 5, and the 2, [which is the] remainder is of the same type as the first, i.e. fifths of a fifth, so we write it beneth the second 5; like this:

If this 3 that is beneath the first divisor was the same as it or greater, i.e. if it was 5 or more, the result would have been an integer, for every five fifths are one integer, so we should divide them by 5; the result of division would have been an integer and the remainder would have been fifths as at first.

Since it is smaller than the divisor, i.e. it is less than 5, which is the first divisor, there is no integer at all and we should not divide [further], as we already achieved our goal, that is, when we divide 70 by 100, each one of them receives 3-fifths, 2-fifths of a fifth, and 2-quarters of a fifth of a fifth.

The rule that follows from this discussion is that if 70 is divided by 100, each one receives 70 parts of 100 parts of the whole.

Since one hundred consists of these three numbers, i.e. from 5, 5, and 4. This is because when we multiply the one by the other and the product by the remaining, the result is 100, meaning the product of 5 by 5 is 25 and when we multiply it by 4, the result is 100.

Since 100 consists of 5 and 4, our saying one part of 100 of the whole is as our saying a quarter of a fifth of a fifth, or our saying a fifth of a fifth of a quarter, because all is the same.

Because, when we say that if 70 is divided by one hundred, each one receives 70 quarters of a fifth of a fifth, we could say 70 fifths of a quarter of a fifth, or 70 fifths of a fifth of a quarter.

This is because when we want to divide the 70 by these divisors, since 70 has a fifth, which is one of these divisors, we should divide it first by 5 and place it last, so that nothing is left to write beneath it. The result of division is 14.

If it would have had a quartet, it would have been appropriate to divide it by 4 and place it before the last.

If a fifth, by 5.

Since it does not have one of them, we arrange the remaining divisors randomly.

For instance, we divide the 14 by 4 and place it before the last; the result of division is 3. Since it is less than the 5, which is the other divisor, we write it beneath it and we write the remaining 2 beneath the 4, which is the second divisor by which we divide. Like this:

Thus, we get more proper fractions, because it is more appropriate to say 3-fifths and two-quarters of a fifth, which is half a fifth, as appears in this diagram, than saying 3-fifths, two-fifths of a fifth, and 2-quarters of a fifth of a fifth.

It is known that the check of this is to convert all of them to the first type, i.e. fifths of a quarter of a fifth, as the last diagram, or quarters of a fifth of a fifth, as the preceding diagram, and this is called "decomposing to a fraction", as will be explained in the second section.

Since it has whole fifths and also quarters of a fifth, we convert all of them first to quarters of a fifth.

It is known that every whole fifth is 4-quarters, meaning 4-quarters of a fifth, as every integer is 4-quarters of the whole.

We get that each one of these 3 whole fifths is 4-quarters of a fifth.

To know how many are they, we multiply 3, which is the number of the fifths, by 4, which is the next divisor; the result is 12. We get that 3-fifths are 12 quarters of a fifth.

We find two beneath it, which is of this type, meaning quarters of fifths. We add it to them; they are 14 quarters of a fifth.

When we want to know how many fifths of a quarter of a fifth are they, we multiply them by 5, which is the divisor that follows them; the total result is 70.

If there was anything beneath this divisor, it would have been also fifths of a quarter of a fifth and we would have add it to them.

Since nothing is found beneath it and the divisors are all gone, we have completed our procedure

Since the result is our original number, meaning 70, which is the small number that we wanted to divide by 100, which has these divisors no more and no less, we know that our procedure is right.

• We wish to divide 140 by 100.

• Since the 140, which is the number that we want to divide, has all these divisors, we place whichever we want last and divide by it; by 5, for instance. The result of division is 28 and nothing remains.

• Since it has a quarter, we divide it by 4 and place it before the last. The result of division is 7.

• Since it is greater than the remaining divisor, we divide it by it; the result of division is 1, which is an integer, as the divisors are gone, we write it aside and beneath it we write the remaining 2. Like this:

Hence, when we divide 140 by 100, each one of the one-hundred is 140 parts of one-hundred, which are 140-fifths of a quarter of a fifth, which are 28-quarters of a fifth, which are 7-fifths, which are 1 integer and 2-fifths.

As this dividend is double the former dividend, so the result of division, which is 1 integer and 2-fifths, which are 7-fifths, is double the former result of division, which is 3-fifths, 2-quarters of a fifth and half a fifth.

Because, we decompose the one to fifths by multiplying it by 5, which is the denominator of the fifths; it is 5.

We add to it the 2 that is beneath it, which are also fifths.

When we want to decompose them to quarters of a fifth, we multiply them by 4; they are 28 quarters of a fifth, and since there is nothing beneath them, we do not add to them anything.

We decompose them to fifths of a quarter of a fifth by multiplying them by 5; the result is 140 fifths of a quarter of a fifth. The divisors are all gone and there is nothing beneath them to add to the result.

We get the same result as our original number.

Chapter Five: Proportions

Rule of Three

• For example, the ratio of 5, which number has the same ratio?

• Example: For the ratio that 5 is to 7, 10 is in the same ratio to which number?

• Or, which number has the same ratio to 14?

• If you say: to which number does 10 has this ratio? What we relate to it is missing, which is the second of the others.

• If we say: which number has this ratio to 14? the related is missing, which is the first of the others.

• When you wish to know the unknown:

• If you have the first of the two that are last, but the second of those that are last is unknown: we multiply the first [of those that are last] by the second of the first two that are known, then divide by that which remains of the three that are known, i.e. by the first of those that are first; the result of division is the unknown.

• If you have the second of those that are last, but the first is unknown: multiply the second [of those that are last], which is known, by the first of those that are first, then divide the product by the second of those that are first; and the result of division is the required unknown.

• Example: if we say: the ratio of 3 to 7 - 5 has this ratio to whom?

We know the related of those that are last, which is called first, it is 5. We multiply it by the second of those that are first, which is 7; the result is 35. We divide it by what remains of those that are known, which is 3; the result of division is 11 integers and 2-thirds and this is the required unknown.

Meaning: the ratio that 3 is to 7 is that same ratio that 5 is to 11 and 2-thirds.

It is clear that as 7 is twice 3 plus its third, which is 1, so is 11 and 2-thirds twice 5 plus its third, which is 1 and 2-thirds.

כן י"א וב' שלישיות הוא כפל ה' ועוד שלישיתו שהוא א' וב' שלישיות

• If 3 golden dinar are worth 50 silver dinar, how many silver dinar will 11 golden dinar be worth?

• If 3 golden dinar are worth 50 silver dinar, how many [golden dinar] will 183⅓ silver dinar be worth?

• The first problem:

• The second problem:

Proportional Triad

• When the mean and one of the others are known, the unknown is extracted, for we multiply the mean by itself , then divide [the product] by the other that is known and the result is the unknown.

• When the two [extremes] are known, the mean is extracted, by multiplying the two that are known [one by the other] and the product is as the product of the mean by itself, i.e. as its square.

The mean is the root - we extract the root of this number by finding a number whose product by itself is as this product and the resulting root is the unknown mean.

• 

• As we say: the ratio of 2 to 4 is as the ratio of 4 to 8

• 4 is the mean, instead of the two [means], which is the second of those that are first and the first of those that are last.

• If one of the extremes is unknown, such as 2, whereas 4 and 8 are known:

• Meaning that one asks: Who has the ratio to 4, as the ratio of 4 to 8?

• Or if he says: How many golden dinar are worth 4 silver dinar, if 4 golden dinar are worth [8] silver dinar?

We know that the product of 4 by 4, which is the mean, that is 16 is as the product of the known 8 by the unknown

Thus, we divide it by 8 and the result of division, which is 2, is the unknown.

• If 2 and 4 are known, and 8 is unknown:

• That is, one asks: the ratio of 2 to 4 - 4 has this ratio to whom?

• Or, if he asks: If 2 golden dinar are worth 4 silver dinar, how many silver dinar will 4 golden dinar be worth?

We multiply 4 by itself; the result is 16. We divide it by [2]; the result of division is [8] and this is the unknown.

• If the unknown is 4, which is the mean that stands instead of the two [means], whereas 2 and 8, the first and the last, are known:

We multiply 2 by 8; the result is 16 and it is the product of the unknown mean by itself, as we have explained; the 16 is the square of the mean.

The mean is its root - the root of 16 is 4.

Chapter Six: Roots

written extraction of roots

description of the procedure

• If it is odd, consider the last digit [= leftmost digit] as if it is units, [then seek] a number such that when we multiply it by itself that product is the last digit or as close to it as possible and we write it beneath it.

If something remains from the last upper digit after subtracting the product of the number that you wrote beneath it by itself, write this remainder above the last digit.

• If the number of the ranks of the number, whose root you wish to know, is even, consider the last digit as tens and what you find in the preceding rank as units.

Then seek a number whose square is the same as these tens and units or as great as possible to be subtract from them and write this number you find beneath the rank that precedes the last rank.

If tens remain after subtracting the square of the number you find from the units and tens that are found in the last two ranks, write them above the last digit; if units [remain], write them above [the digit] that precedes the last digit.

• If no tens result from this doubling, place the units of the double [the number] beneath the rank that precedes the rank in which you placed [that number] at first.

• If tens or more result, place the tens beneath the rank in which the number was at the beginning, and the units in the rank that precedes it.

• If there are no units there, ?

Write it in the mentioned rank, i.e. in the rank that precedes the ranks in which you wrote the double of the first number.

Multiply it by the first numbers, except the the one you wrote first that is crossed by a pen.

Subtract the product from the digits that above them.

Multiply [the number] by itself and subtract [the product] from the rank above it.

Wherever there is a remainder, write it above the digit from which it remains.

• When you double the number and write its double in the preceding rank, if the digit that is above it is not enough to subtract [the double] even once, and that enough remains in the preceding rank to subtract from it the product of the other digit by itself, write 0 before it and lower it by one rank, whether there is a 0 there or any other digit. Then seek a number to write before it, so that you multiply it by each of the [digits of the root] and by itself and subtract every [product] from what is above it.

Write the remainder above the digit that is above it and lower again the [digits of the root] by one rank, as long as you lower them, whenever you lower them, as they are, without doubling, except for the last digit that is generated at that same phase that you double it.

• If no tens are generated [in double the digit of the root], write it in the rank that precedes the ranks in which you write the other digits as they are lowered.

• If a ten is generated, add it to the digit that you write first to the right, and if this digit is only 0, write the 1, i.e. the ten instead of it. Write the units that are generated with the ten from this double, in the preceding rank.

• If no tens are generated [in double the digit of the root], as when the last digit is a five and its double is a ten without units, write the ten as I said instead of the 0, or add it to what you find in the rank to the right, and since there are no units, write 0 before these ranks.

examples

• We wish to seek the root of 344680129066.