Difference between revisions of "Excerpts Attributed to Immanuel Bonfils"

Introduction - Decimal Fractions

Another Version - MS Paris 903

MS Paris 1081 15r-16v

Table – multiplication of sexagesimal fractions (Firenze)

Table – multiplication of sexagesimal fractions (Firenze)

Multiplication of integers (Firenze)

• Example for the multiplication methods: we wish to multiply 127 by 355.

We write 127 on the upper line and 355 on the other lower [line]. This is its diagram:

We multiply 7 by 5; it is 35.

We write 5 in the first rank, corresponding to 7; and keep the 3, in order to add it to the second rank.

We multiply 7 by the second lower 5; the result is 35.

We add the 5 to the 3 we kept; the result is 8.

We write 8 in the second rank, beneath the 2 of the upper line and we write the 3 in the third rank, beneath the 1 of the upper line, with the result of multiplication of that rank.

We multiply the first 7 by the lower 3; the result is 21.

We add the 1 to the 3 we kept; it is 84.

We write 4 beneath the 1 in the third rank and 2 in the fourth rank.

We multiply the upper middle 2 by the first 5 on the lower line; the result is ten.

We write a zero in the second [rank] and 1 in the third [rank], beneath the 3 of the lower line.

We multiply the upper 2 by the second upper 5; the result is ten.

We add the 10 to the 1 we kept; it is 11.

We write 1 in the third [rank] and keep the 1 in the fourth [rank].

We multiply 2 by 3; it is six.

We add the 6 to the 1; it is 7. We write it in the fourth [rank].

We multiply 1 by the first 5 in the lower line; we write 5 in the third [rank].

עוד כפלנו א' על ה' הראשון שבטור השפל וכתבנו ה' בשלישית

We multiply 1 by the second 5; we write 5 in the fourth [rank].

We multiply 1 by 3; we write 3 in the fifth [rank].

The multiplication is complete.

We sum up the result of multiplication of the two rows:

We write 5 in the first rank.

8 in the second.

4, 1, 5 are ten; we write 0 in the third and mark a point in the fourth to sign the ten.

We sum up 2, 7, 5; it is 14. Then the reserved; it is 15. We write 5 in the fourth rank and a point in the fifth.

We add one to the 3; it is 4. We write 4 in the fifth rank.

The addition is complete.

We consider the [digits] in the upper row as if they were units; they are ten according to this way.

We cast out the nine; 1 remains. We write it next to the upper row.

Then, we calculate the bottom row; we find the result is 13.

We subtract the 9; 4 remains. We write it next to the one.

Then we multiply 1 by 4; it is 4. We keep it.

We calculate the result, which is 45085; the result is 22.

We cast out 18, which is removed by nines; 4 remains and it is equal to the product of the scales of the two rows. Therefore we know that our calculation is correct.

As if there are 3 and 4 in the upper [row], then the result is 7; and in the bottom [row] there are 2 and 3, so the result is 5: multiply 7 by 5; the result is 35. The remainder from the nines you cast out is 8.

Division of integers

You shift the whole number backwards, i.e. the digit 8 to the 9 before it and take ten for each unit, i.e. with the 9 before the 8, they are 89. See how many times the 9, which is the last digit of the bottom row, is in 89. But, keep in mind that the digit 4, which is before the 9 in the bottom row, should be as many times in the digit 4, which is before the 9 in the upper row. As if you say that you can take 9 times 9, which is 81, from the number 89 and 8 remains. With the preceding 4, it is 84. So, you can take 9 times 4 from the number 84.

Therefore, take 9 and write it first in the middle row, as far from the digit, from which you took the quotient, which is 9, as the number [of ranks] that the last digit of the bottom row, which is 9, is far from the rank [of units], which is four ranks. So, write the 9 you receive from the division in the rank of tens, which is beneath the 2 of the upper row.

An explanatory supplement that does not belong to this section:

Extracting roots

A hundred, which is analogous to one, whose root is ten, which is analogous to one.

Four hundreds, which is analogous to four units, whose root is twenty, which is analogous to two.

9 hundreds, which is analogous to 9 units, whose root is thirty, which is analogous to three.

Like one thousand and six hundred, which is analogous to sixteen, whose root is forty, which is analogous to four.

Two thousands and 500, which is analogous to 25, whose root is fifty, which is analogous to five.

Three thousand and six hundred, which is analogous to 36, whose root is sixty, which is analogous to six.

Four thousand and nine hundred, which is analogous to 49, whose root is seventy, which is analogous to seven.

Six thousand and four hundred, which is analogous to 64, whose root is eighty, which is analogous to eight.

Eight thousand and one hundred, which is analogous to 81, whose root is ninety, which is analogous to nine.

• As, if you say, for example: we have the number 4 in the seventh rank, which is an odd rank, like in this diagram:

4 0 0 0 0 0 0       2

Its root is 2. As we explained, 2 should be written in the fourth rank, which is the middle rank that has 3 rank before it and the same after it.

• Suppose first that it has five ranks, for example: 12345.

0 0 3 2 0 1 4     1         5 4 3 2 1     2 3 3     4 6

4 0 0     1 2 9 0     1 3 8 9

5 4 3 2 1

The last rank is odd, so it follows the rule of the units. Therefore, we consider the last digit 5 as if it is 5 units. The closest preceding root is 2, which is the root of 4.

Therefore, we write 2 in the middle, beneath the digit 3, and we draw a line beneath the 2 with some space, so that we can multiply the resulting digit of the root.

We multiply the digit 2, which is the root, by itself. The product should be written following it, as far from it as it is far from the units. Zeros should be written as the number of the ranks before it, starting straightly corresponding to the 2 in its rank, as we say now corresponding to the 2, we start writing 2 zeros, as the digit 2 is far from the units by two ranks. After the 2 zeros, we multiply the 2 by itself. Understand and remember it for every product. Then, subtract the product from the first corresponding digit, as known in the subtraction procedure, and the remainder remains above the upper row.

Then, multiply the root by two, i.e.: if it is 3, write 6 beneath it; if it is 4, [write] 8. In this example, it is 2, so write 4 and this is the root multiplied by two.

Divide the remainder of the first digit by this 4: take as much as you can, so that you can take the square of the quotient. As you say in this example: we return the 1 back, because we cannot take 4 from it even once. We return it back to the 4 that precedes it; it is 14. We say: how many times can we take 4 from 14? We take 3, because we can take 3 times 3, which is the square of the quotient, from what is left before. We write 3 before the first root.

We multiply the 3 first by itself and then by the doubled root.

We start with the 1 next to it: write as many zeros as the distance of the digit 3 from the units; it is one zero corresponding to the 3.

We say: 3 times 3 is 9.

We say: 3 times 4 is 12. Write 12.

We subtract the multiplied row from the the remainder above, each from the one that corresponds to it, according to the rule of subtraction, and what is left remains above in the upper row.

Then, multiply the second digit of the root, which is 3, by two. We write 6 beneath it.

We divide again all that remains above by the whole doubled root, which is 46.

We take as much as we can, so that we can take the square of the quotient from the last [digit]. As you say in this example: we return the 1 back, because we cannot take 4 from it; it is 14. We take 13, because we can take 3 times from the remainder and we can take 3 times 3 from the remainder after it. We write 3 before the 6, which is in the rank of the units.

First, we multiply this 3 by itself; it is 9. We write it corresponding to it, because the digit 3 is in the rank of the units.

Next, we multiply 3 by 6 and 3 by 4. We subtract the products from what is left above and the remainder remains above.

The remainder above is 32 and the root is 233.

The explanation is complete.

• The first rule: when you extract the first root from the last digit you have, if it is in an odd rank, or in the preceding to the last, if it is in an even rank, you should write that root in the middle rank, as I have shown you and take as much as you can. If you cannot take anything from what remains in that last rank, and you take from the [rank] that is second to it, write it as far from the place, from which you take the quotient, as the first root is far from the units. So, you should write a zero in the place that precedes the first root, and before this zero you should write the result of the division.

• The second rule: when you divide again what remains of the number you have by double the root, do not take the last number from which you took it first, even if you see that it is possible to take from it, because when you take the root from it first, you already take from it everything that should [be taken from it], as you see in our example

6     3 9 9   1     2

• We wish to know the approximate root of 399.

We take the closest root, which is 1, and we write it in the middle rank, which is beneath the second 9.

We subtract 1 from the last digit, which is 3; 2 remains above the 3.

We double the digit 1; it is 2.

We do not say and look how many times the 2, which is double the root, is in the number that remained from the first number, which is 2, as we can take 2 from 2 once and then we can also take the square of 1 from the remainder. But this is not appropriate to do, because it would be appropriate to write the resulting 1 far from the number, from which we took the quotient, when the digit 1 written first is written according to this rule. Since we have already taken all that is needed from that rank, so we return the 2 back on the 9 that precedes it and it is 29.

We say: how many times 2, which is double the root, is in the number 29, so we can take the square of the quotient from the preceding rank.

This is enough for this rule and it follows the third rule.

• The third rule: when you divide again the remaining number by double the root, do not bother yourself to see if you can take ten times the divisor from the dividend, as stated in the example above: we see how many times 2 is in the number 29 and we say that it is ten, because this is impossible, even if it is more than ten, because you will not be able to take ten times ten from the preceding rank. It is not appropriate to bother with it, and this is also the rule of this matter.

• The fourth rule: as we said in the second rule that when we divide the remainder again by double the root, we do not take again from the rank we took first, this is when we double the root and we do not add a rank to it, as if we double 2, it is 4 in the same rank, the same for 3 and 4. But, if the root is 5, or 6, or 7, or 8, or 9, when we double the root, one rank is added to the root. As if you say: the root is 6 tens, so double the root is 2 tens and one hundred, then one rank is added to the root. According to this, when you divide again, you take from the rank, from which you took first, and you write it also far from the last digit as the beginning of the doubled root is far from the units. This is what we wanted to clarify.

Another Version - MS Paris 903

הן משלמי' לבד או נשברי' לבד הן ממספר שלמי' ונשברי' יחד

As in the hundreds: 100, whose root is 10; 400, whose root is 20; 900, whose root is 30, and so on.

As the thousands, in which there are six squares.

As in the thousands:

One thousand and 600, which is analogous to 16, whose root is 40, which is analogous to 4.

Two thousand and 500, which is analogous to 25, whose root is 50, which is analogous to 5.

3 thousand and 600, which is analogous to 36, whose root is 60, which is analogous to 6.

4 thousand and 900, which is analogous to 49, whose root is 70, which is analogous to 7.

6 thousand and 400, which is analogous to 64, whose root is 80, which is analogous to 8.

8 thousand and 100, which is analogous to 81, whose root is 90, which is analogous to 9.

• Example: we find 4 in the ninth rank, like this:

4 0 0 0 0 0 0 0 0         2

We know that this number is square, because it is in an odd rank, so it is analogous to the units; in the units 4 is a square and its root is 2, so its root is also 2.

The position of the 2 should be in the middle between the rank of the units and the rank of the square, meaning the fifth [rank].

As if you say: we find 36 in the eighth rank, which is even, and it is analogous to 36, which is a square in the tens and its root is 6, so its root is also 6 and the 6 must be written between the beginning of the square number and the units, in the middle, that is, in the fourth rank, which is 3 ranks away from the units, and so is its distance from the [seventh] rank, where the square number begins.

3 6 0 0 0 0 0 0       6

• Below is an example of 5 ranks as follows:

0 0 3 2 0 1 4     1         5 4 3 2 1     2 3 3     4 6

4 0 0     1 2 9 0     1 3 8 9

5 4 2 8 9       3 2

5 4 3 2 1

You know from the above that this number [is in] an odd [rank] and it is analogous to the units, i.e. the 5. The closest root to it is 2, which is the root of 4, which is closest to it, preceding it.

The place of this 2 as above is in the third rank beneath the 3.

Now, see how many ranks there are from the units to the place of the 2, which is your root, and start writing zeros from there, as the number of these ranks.

As, if you say: from the units to the 2 there are 2 ranks, so we write two zeros beneath the line, starting from the 2 itself and going to the left.

Then, we write the product of 2 by 2 there and this is 4.

Subtract the 4 from the 5 of the required number; 1 remains.

Write 1 above the 5 and strike out the 5, because it is not needed any more.

Now, you should to know the root that is closer than the 2:

You should take a number to add to the root, so that you add double that number to twice the root, which is 2 in our example, in the rank of hundreds, as well as the product of that number by itself.

The proof of this is that we assume a square number of 4 tens of thousands, so its root is 2 hundred, as you have seen, and its root is one side of the square itself. Because a length of 2 hundred cubits by a width of 2 hundred cubits is 4 tens of thousands of cubits by cubit.

Suppose we add thirty to the root of this square, so one side of this square is 230.

We find that when you put 30 strips, each of which is two hundred cubits long and 1 cubit wide, that is, one strip of two hundred by 30, and you add them to one side of this square - as, if you say to its east - then, its length exceeds its width - its length is 230 and its width is 200.

Then, you add another strip of 200 by 30 to the south of the square, so its width is also 230.

Therefore, you already added to this [square] a product of 30 by 200 - once to the east and once to the south.

This is a product of 30 by 400.

There is still a "remote corner" missing in this rectangle, and it is the southeastern corner, which is not filled until after you put there a square of 30 by 30. Then, the rectangle becomes an equilateral square, whose root is 230.

Now, divide the upper number by the 4 and see how many times 4 is in the last 1 with the 4 that precedes it, when you shift it backwards and they become 14:

It is found 3 times, but you need to see if it is enough to fill the "remote corner", which is 9 and it is the product of 30 by 30.

If it was not enough, I would take only 2, but in our example we can take 3.

We write 3 before the 2, to its right.

Again do as above, like this:

See how many ranks there are from the units to the 3; there is one.

So, write one zero beneath the line in the second rank, i.e. beneath that 3.

Then, multiply 3 by itself; the result is 9.

Write it in its place beneath the line, to the left of the single zero you wrote below the 3.

Multiply again the 3 by the 4, which is twice the root; the result is 12.

Write it in its place after the 9, beneath the line. Because the 3 is of the rank of tens and the 4 is of the rank of hundreds and the product of tens by hundreds is thousands. So, start writing the 2 beneath the rank of thousands and the 1 beneath the tens of thousands.

Then, subtract it from the upper number, each from its corresponding:

Meaning, the 1 from 1 - nothing remains; the 2 from 4 - 2 remains.

Since we cannot subtract the 9 from 3, we shift 1 backwards and write 1.

We subtract the 9 from the 1 that is shifted backwards to the 3, so they are 13; 4 remains.

We write 4 and we write all above the upper line each in its rank as you see in our diagram.

Strike out the 4, 3 and 1, because they are not needed.

You find the approximate root 230.

You want to increase the square and root further, because there are more digits of the original required number, since you are still left with 1421.

Multiply the 3 also by 2; it is 6.

Write it beneath the 3.

Now, divide the number remaining from the required by double the whole root, which is 46, meaning 460, because the root is 230.

See how many times 4 is in 14; it is 3 times in it.

Write 3 before the 23 of the root, beneath the 1.

Multiply 3 by itself; it is 9.

Write it beneath the line and beneath the 3 of the units, in order to complete the "remote corner", and since the 3 is not at all far from the rank of units, there is no need to write a zero.

Multiply the 3 also by twice the root, which is 46; you find 138.

Write it in its place, meaning, start from the lower 9.

The result of multiplying 3 by itself and by double the root is 1389.

Subtract this number from what you have left of the required number, which is 1421:

Subtract 1 from 1; nothing remains.

Subtract 3 from 4; 1 remains.

Since we cannot subtract 8 from 2, we should shift this 1 backwards, so nothing remains in this rank also.

Subtract 8 from 12; 4 remains.

We still need to shift 1 backwards from the 4; 3 remains. Write it down.

Subtract 9 from 11; 2 remains.

Write it in its place and crossed out 1421 because it is no longer needed.

So, the required approximate root of 54321 is 233.

The square of 233 is as the sum of all the squares of the roots you extracted, which is 54289 and the remainder is 32.

If you add 32 to the square, you receive the required number.

Division of sexagesimal fractions

Extracting roots of sexagesimal fractions

Multiplication of sexagesimal fractions

A Rule for Checking the Squares

MS London 7r-v

1) We multiply the result of the division according to the multiplication method by the bottom row, which is the divisor, then we add the remainder of the division to the product, if there is anything left. If the sum is equal to the upper row, which is the dividend, we calculated correctly.

2) We multiply the scales of the result of division by the scales of the bottom row; we take the result. If there is anything left, we add it to it. If it equals the scales of the upper row, our calculation is correct.

Chapter on Summation

• As if you say: we wish to know the sum from one to ten, as 1 with 2 - it is 3; 3 with 4 - it is 7; and you want to know easily:

If their number is even, as ten: we add 1 to it. Take a half of ten, which is five, and multiply the five by 11 and this is the last number [= the sum].

If [their number] is even, as from one to seven: add one to seven; it is 8. Take half of this number; it is 4. Multiply 4 by 7 and this is the required.

• If you wish to find an unknown term, as if you say: the ratio of 4 to 6 is the same as the ratio of 8 to the unknown and you wish to find [the unknown term]

[Multiply] the means [by each other], 6 by 8; it is 48.

Divide [it by] the first number, which is 4; you receive [12], which is the unknown.

• Example: one asks: I added up numbers and the result is 465; how much is the last number?

We double 465; the result is 930. We know that the preceding square is 9 hundred and its root is 30. Between the square and the double there is none but 30, so you understand that the calculation is correct and the required is the root.

Word Problems

"How Many" Problem

• Question: a man passed by a group of people.

He said to them: hello one hundred people.

They answered him: we are not one hundred people, but all of us, and other like us, and half of us, and a quarter of us with you will make one hundred.

We take one for them; add one to them; add its half to them; it is two and a half. We add its quarter; they are 3-quarter. Since they are quarters added to integers, we convert the integers into quarters; [2] and 3-quarters are 11.

Since he said that they are [100], their number without the addition of the one who spoke with them is 99.

We convert the 99 into quarters; they are 396.

We divide them by 11; it is 36 and this is their number.

"If You Give Me" Problem

• Question: four buy one item.

One said to his friends: each one will give half of what is in his purse and I will give everything I have in my purse and I will buy the item.

The second [said]: each one will give a third of what is in his purse and I [will give] everything I have in my purse and I will buy the item.

The third said: each one will give a quarter and I [will give] everything I have in my purse and I will buy the item.

The fourth said: each one will give a fifth and I [will give] everything I have in my purse and I will buy the item.

How much is the price of the item and how much does each have in his purse?

אחד אמר לחבירו כל אחד יתן החצי אשר בכיסו ואני אתן כל מה שיש בכיסי ואקנה החפץ
והשיני כל אחד יתן שליש כל מה שבכיסו ואני כל מה בכיסי ואקנה החפץ
והשלישי אמר כל אחד יתן הרביע ואני כל מה שבכיסי ואקנה החפץ
והרביעי אמר כל אחד יתן חומש ואני כל מה שבכיסי ואקנה החפץ
כמה ערך החפץ וכמה לכל אחד בכיסו

The price of the item is 46 dinar.

The first has 15 pešiṭim.

The second has 23 dinar and 9 pešiṭim.

The third has 31 dinar and 3 pešiṭim.

The fourth has 35 dinar.

It is also possible to reach a half, a third and a quarter: the item 17 pešiṭim; the first - 5; the second - 11; and the third 13.

"Find the Amount" Problem

• Question: A man sells 13 measures for 23 pešiṭim.

How many measures will he sell for 7 pešiṭim?

כמה מדות יתן בז' פש‫'

As the ratio of 7 to 23 so is the ratio of the unknown to 13.

We multiply the extremes that are known; they are 91.

We divide them by 23; the result is 3 measures and 22 parts of 23 of one measure.

• We reverse the matter: we want to know for how much will he sell 7 measures?

We do like this: we multiply the extremes; they are 161.

We divide them by 1[3]; they are 12 pešiṭim and 5 parts of 13 of one pašiṭ.

If it is asked for how much will he sell 7 measures, the ratio of the unknown to 23 is the same as the ratio of 7 to 13.

Pursuit Problem

• Question: A man walks 29 miles a day.

After 10 days, another starts walking 37 miles a day.

When will he catch up with him?

אחר י' ימים נסע השני ההולך בכל יום ל"ז מילין
מתי ישיגהו

Multiply the miles he walks by 10 days; they are 290.

We divide it by the difference between the two velocities, which is 8; it is 36 days and a quarter of a day.

As the ratio of 10 days to the unknown so is the ratio of the difference, which is 8, to 29.

Encounter Problem

• Question: In how many days a 100 [miles] walk will be completed by two walkers at a [speed of] 19 and 17 towards each other?

It is easy to know that it is as the ratio of the sum of both to 100. Divide it by it; it is 2 and 28 [parts] of 36 [...]; 28 of 36 are 7-ninths of a day.

If you want to know how many hours they are: it is known that the hours of the day are 12.

[As the ratio of] 7 to 9 so is the ratio of the unknown to 19.

Payment Problem

• Question: A man had 3 [workers].

[He said] that he would give the first 5 zehuvim and he would work for him for 20 days, but he did not want to finish the whole work.

He said to the second that he would give him 4, but he did not want to finish [the work].

He said to the third that he would give him 5, but he did not want to finish either.

The three of them went and did everything together and the employer paid each one equally.

ואמר לשיני שיתן לו ד' ולא רצה לגמור
אמר לשלישי שיתן לו ה' ולא רצה ג"כ לגמור אמר לשלישי שיתן לו ה' ולא רצה
והלכו שלשתן ועשו הכל ביניהם והשוכר נתן לאחד אחד שוה

This is similar to the question about the moneychanger, but we define the zehuvim divided among the workers as the coins, the hired work is the zahuv, the common denominator is 60 and the sum of the parts is 47.

As the ratio of 47 to 60 so is the ratio of the zahuv to the unknown.

The part is worth one zahuv and 13 [parts] of 47 of one zahuv.

We wish to know: how much is the share of each:

The owner of the 3: 6 days and 2-thirds of a day [for one zahuv].

The owner of the 4: 5 days for one zahuv.

The owner of the 5: 4 days for one zahuv.

The 3: divide his share by 47. As the ratio of 13 to 47 so is the ratio of the unknown to 20, which is the number of hours he works for one zahuv.

The 4: as the ratio of 13 to 47 so is the ratio of the unknown to [15].

If you add up all, the resulting sum is 20 days.

Boiling Problems

Question: A man wanted to cook ten measures of must so that one-third remains. After 2 [measures] were reduced and 8 remained, 2 [measures] overflowed, so they were 6. Thus, one asks: how much should remain, so that it will be cooked according to the first plan, when it is known that at the beginning 3 and one-third should have been reduced?

As the ratio of 6 to 8 so is the ratio of the unknown to 3 and a third.

Whole from Parts Problems - lance

• Question: a lance, a third of it is in the water, a quarter of it is in the soil, and [up above the water] it is 7 cubits [long].

What is the length of the whole [lance]?

כמה גבהות כלו

The common denominator is 12.

We subtract its parts from it; [the remainder] is 5.

As the ratio of 5 to 12 so is the ratio of 7 to the unknown.

Multiple Quantities Problems

• Question: A man ordered to give 20 pešuṭim to 20 people - men and women and children - he gave 2 pešiṭim to each man, a half to each woman, and [a quarter to each child].

How many [men, women, and children were there]?

כמה לאיש

For 8 men - 16 pešiṭim.

For 8 children - 2 pešiṭim

For 4 women - 2 pešiṭim.

• Question: A man said to his slave: Take 30 pešiṭim and buy me 30 poultry - a goose for 2 pešiṭim, a pheasant for a half, and a pullet for one pašiṭ.

How many were there of each kind?

כמה יהיו מכל מין

6 pullets, 8 geese, and 16 pheasants.

Whole from Parts Problems - fish

• Question: a fish is 15 fingerbreadth [etzba'ot] in its body, without its tail and its head; its head is one-third of the whole fish and its tail is its quarter. How much is its whole with the head and the tail?

36 fingers: its third is 12; its quarter is 9.

[?] you find that it is so.

• Question: A thief walks a fixed distance and the pursuer [walks] on the first day 1, on the second day 2 and so on he adds 2 every day, so the distance increases twice [the distance of] the thief.

In how many days will he catch the thief walking the entire distance he walked each day?

• Question: Shimon earned a seventh of his money on one day, the next day he earned a fifth of the whole amount, on the third day a third of the whole amount and the total sum is 10 dinar.

We ask: how much is the amount of money?

ונשאל כמה הממון

The common denominator is 105.

• Question: to weigh 60 with 11 stones; also [to pay] the rent for 4 years and 2 months with 18 rings

The result for the weight is with doubles.

Do for rings as follows: 1, 2, 4, 8, 16.

Duplicate until the number is obtained.

For stones the result is with triples: 1, 3, 9, 27

Glosses on Abraham Ibn Ezra’s Book of the number (P1026; London)

The author of the Sefer Yezira

Sefar

The writing: through the faculties of the letters that are indicating him, the creation is done by the letters.

Sefer

It is the number, because composition and merging are only done by that one number is the duplication of the other.

Sipur

Because the scholars of this science assumed that they were leading down a spiritual force that gave power to that composition, so that the action of this power is not through anything, but not through anything is possible only through God.

Since

Higher decade

The remainder is the required

Explanation: a third multiplied by a third is a ninth.

Therefore, when we multiply a third of a number by its third, we receive a ninth of the required square.

When we take the same in the higher rank, we receive ten times the ninth of the square.

Therefore, when we subtract from it the square of the third, which is a ninth of the square, the remainder equals the required square.

The sum is the required

• Such as 12 and 18.

We multiply a third of the one, which is 4, by the third of the other, which is 6; the result is 24.

We rise it one rank higher; it is 240.

Subtract the product of the two-thirds, which is 24; 216 remains and this is the product of 18 by 12.

• If they [are not divisible by three], but exceed by one:

• As 13 and 19, we do as follows:

We calculate the product of 12 by 18; the result is 216.

Since each of the numbers exceed by one over [the divisible by three], we add 12, 18, and one; the result is 247 and this is the product of 13 by 19.

• If the two numbers are one less [than the divisible by three]:

• As 11 and 17:

We calculate the product of 12 by 18; the result is 216.

We subtract 11, 17 and 1 from 216 for the deficiency; 187 remains and this is the product of 11 by 17.

• If one exceeds by one and the other is less by one:

As 11 and 19, or 13 and 17.

We calculate the product of 12 by 18; the result is 216.

• If they are 11 and 19:

The [deficient] is 11 and the [exceeding] is 1[9], so the [exceeding] is greater than the [deficient] by [8; minus 1; it is] 7. Subtract it from 216; 209 remains and this is the required.

• If they are 13 and 17:

The [deficient] is 17 and the [exceeding] is 13, so the excess of the greater is [4; plus 1; it is] 5. We add it to 216; it is 221 and this is the required.

• If they are 18 and 13:

We add 18 to 216; it is 23[4] and this is the required.

• If they are 18 and 11:

We subtract 18 from 216; 198 remains.

Example: we wish to multiply 29 by 31.

Explanation: another example: when we wish to multiply, for instance, 8 by 12, whose distance from 10, which is a decade, is the same, which is 2.

We multiply 10 by itself; it is 100.

We subtract 4 from it, which is the square of 2.

The explanation of this is that the multiplication of 8 by 12 is the multiplication of 8 by 10 and by 2.

The multiplication of 10 by 10 consists of the multiplication of 10 by 8 and by 2.

The multiplication of 10 by 2 consists of the multiplication of 8 by 2 and 2 by 2.

We find that the excess of 10 by 10 over 8 by 12 is 2 by 2.

From this you understand all the other examples, and this is according to proposition 6 of Euclid's Book II.

The rule is the same for any two different numbers, even if there are no decades: we add to the smaller number half the difference between the two numbers, square [the sum], subtract from the square the square of half the difference and the remainder is equal to the product of one of the two numbers by the other.

But this scholar wanted to make it easier, not harder, and for this reason he explained it only for the numbers whose distance from a decade is the same.

Know that if there are two digits to multiply

Explanation: if you wish to multiply 2 by 40, it is just a multiplication of one rank by another, but if we have one digit by two digits, or two by two, the ranks must be multiplied by each other according to the numerical value of the digits.

Explanation: he said it in actu, although sometimes it can be done with less, because he said above: if they are two by two, it should be done 4 times, and then he explained that it can be done with less, so here too it can be done with less. Understand.

Does not assume any change

One is like a point

Explanation: because the point is like the foundation of the circle because the one who draws a circle, before drawing it, draws a center, on which he establishes the circumference of the circle.

The one is the foundation of the number, and therefore it cannot be divided like the point, since it is one.

But since every sum has the name of one, being composed of many individuals, like the body, which is composed of four phlegms and other organs, which have homogeneous parts, and the hand and the arm, and the rest of the organs, hence it is undoubtedly possible for it to be decomposed into parts, except for the intellect, in which the speech is created, for this reason it was decided that the simple division is done in the intellect, because without the intellect it is impossible in any way.

Another explanation: all this is an explanation for dividing the one into fractions, and just as the point is an indivisible thing and all lines are drawn from it, so is the one, and for this reason it cannot be divided.

Consists of surfaces

Explanation: he did not mean that the body is a chemical composition of surfaces and not of lines and points, as explained in the physics books, but in his statement he meant the surfaces of the encompassing objects and many in the human body, such as the skull that is encompassed by one surface, and so are many organs, and the body is called one.

It is similar to an odd number

He did not really need to take sixths, because the thirds were enough for him: by multiplying [3] by 5; it is 15. We multiply it by 7; it is 105. We multiply it by 8; it is 840 and this is the common denominator.

We look for the first number which is 2-thirds of a quarter of a fifth; it is 28.

Because the fifth is 168.

Its quarter is 42.

Its two-thirds are 28 and this is the first number.

We know that its eighth is 105.

Its seventh is 15.

Its six-sevenths are 90.

We multiply 28 by 90; the result is two thousand and 5[2]0.

We divide it by 840, which is the denominator; the result is 3 and they are parts, of which 840 are one unit, which are a fifth of a seventh of an eighth, the eighth [of 840] is 105, its seventh is 15 and its fifth is 3 and this is the required.

This is clear, because the square of the entire line, which is 16, is equal to the two squares of 12 and 4 in the proposition 4 of Euclid's II.

The product of ​​6 by 8 is equal to the surface enclosed by 12 and 4.

Since the ratio of 4 to 6 is the same as the ratio of 8 to 12.

Therefore, the square of 16 is equal to the two squares of 12 and 4 and double the surface enclosed by 6 and 8.

But [the sum of] the two squares of 6 and 8 is equal to the double the surface enclosed by 6 and 8, with the square of [the difference] between 6 and 8, as explained in II.5.

Therefore, the square of 16 with the square of the difference between 6 and 8 is equal to the four squares of 4, 6, 8 and 12. This is explained from the second and third.

This is clear from 19.V: when the sum to the sum is as the ratio of the whole to the whole, then the ratio of the remainder to the remainder is as the ratio of the whole to the whole.

Therefore, in his second example, since the ratio of the unknown difference between 3 and 4 to the difference between 4 and the unknown number is as the ratio of the known 3 to the unknown 6, and the unknown is greater than its excess over 4, then 3 is greater than the difference between 3 and 4, according to Euclid's 14.V.

The difference between 3 and 4 is known, so when we subtract it from 3, the remainder is known and it is 2.

We also know that 4 is the remainder, when we subtract from 6 the excess of 6 over it.

Now, we have three known numbers:

The first is 2 and it is the remainder from 3, when the difference between 3 and 4 is subtracted from it.

The second is 3, which is the known whole.

The third is 4, which is the remainder from the unknown, when its excess over 4 is subtracted from it.

We know that the ratio of the subtracted from 3 to the subtracted from the unknown is as the ratio of 3, which is the whole, to the unknown.

Therefore, the ratio of 2, which is the remainder, to 4, which is the remainder, is as the ratio of 3 to the unknown.

We multiply the means, which are 4 and 3; it is 12.

We divide 12 by 2; it is 6, etc.

Understand the others from Euclid's 18.V.

Proposition 16 is derived this way: the first and the seventh are taken, and they are cardinal points. One proposition is derived from both of them. From the fourth and tenth, which are cardinal points, another proposition is derived. Then, from both of them, one proposition is derived, called the 16th, and is the strongest of them all.

I have found, with the help of God, proposition 16: We take the first and seventh propositions and derive one proposition from both of them. We take proposition 15, and from these two, we derive another proposition, which is the 16th, and this is a secret.

• Question: three went to the market to buy a fish, whose price is 5 pešuṭim.

One said: I will give everything in my purse and you will give only half of what is in your purse.

Another said: I will give everything in my purse and you will give only a third of what is in your purse.

Another said: I will give everything in my purse and you will give only a quarter of what is in your purse.

How much is in the purse of each of them?

אמר אחד אני אתן כל מה שבכיסי ואתם לא תתנו רק החצי שבכיסכם
אמר האחר אני אתן כל מה שבכיסי ואתם לא תתנו רק שליש שלכם
אמר האחר אני אתן כל מה שבכיסי ואתם לא תתנו רק הרביע שבכיסכם
כמה יהיה בכיס כל אחד

We look for the common denominator; it is 12.

29: we can extract three numbers from it, so that the three [sums] will be equal to the same number: 5, 11, 13.

The owner of the 5 will ask for a half.

The owner of the 11 will ask for a third.

The owner of the 13 will ask for a quarter.

• Question: Reuven, Shimon, Levi and Yehuda.

Reuven told Shimon: everything in my purse plus a half of what is in your purse is 5 dinar.

Shimon told Levi: everything in my purse plus a third of what is in your purse is 5 dinar.

Levi told Yehuda: everything in my purse plus a quarter of what is in your purse is 5 dinar.

Yehuda, the fourth, told Reuven, the first: everything in my purse plus a fifth of what is in your purse is 5 dinar.

How much is in the purse of each of them?

אמ' ראובן לשמעון כל מה שבכיסי עם חצי שבכיסך הוא ה' די‫'
אמ' שמעון ללוי כל מה שבכיסי עם שליש שבכיסך הוא ה' די‫'
אמ' לוי ליהודה כל מה שבכיסי עם רביע מה שבכיסך הוא ה' די‫'
אמ' יהודה הרביעי לראובן הראשון כל מה שבכיסי עם חומש מה שבכיסך הוא ה' די‫'
כמה בכיס כל אחד

The answer:

Reuven: 37 pešuṭim and 97 parts of 11[9] of one pašuṭ

Shimon: 44 pešiṭim and 44 parts.

Levi: 46 pešuṭim and 106 parts.

Yehuda: 52 pešiṭim and 52 parts.

We find that all that Reuven has with half of what Shimon has is 5.

Also what Shimon has with a third of what Levi has.

Also what Levi has with a quarter of what Yehuda has.

Also what Yehuda has with a fifth of what Reuven has.

We subtract the common half of Shimon; what Reuven has remains equal to half of what Shimon has with a third of what Levi has.

What Shimon has is equal to two-thirds of what Levi has with a quarter of what Yehuda has.

So, what Reuven has is equal to two-thirds of what Levi has with an eighth of what Yehuda has.

What Levi has is equal to 3-quarters of what Yehuda has with a fifth of what Reuven has.

So, what Reuven has is equal to 5-eighths of what Yehuda has with 2 parts of 15 of what Reuven him self has.

Reuven has 75; Shimon has 88; Levi has 93; Yehuda has 104

Or, Reuven has 37 and a half; Shimon has 44; Levi has 46 and a half; Yehuda has 52

And so on for their halves.

How much money did he have?

כמה הוא הממון שבידו

• Example: he spends 48 pešiṭim every day and at the end of three days he has nothing left.

6 is the number that when doubled in a double ratio as many times as the number of days, which is 3, the result is 48.

Because, when we double 6, the result is 12.

When we double 12, the result is 24.

And when we double 24, the result is 48.

We subtract 6 from 48; 42 remains and this is the amount of money he has.

• Or: he spends 18 pešiṭim every day and at the end of five days he has nothing left.

One half and one part of 16, which is 9 parts of 16, is the number that when doubled in a double ratio as many times as the number of days, which is 5, the result is 18.

Because, when we double one half and one part of 16, the result is 1 and one-eighth.

When we double 1 and one-eighth, the result is 2 and a quarter.

When we double 2 and a quarter, the result is 4 and a half.

When we double 4 and a half, the result is 9.

And when we double 9 for the fifth time, the result is 18.

We subtract a half and one part of 16 [from 18]; the result is 17 and 7 parts of 16 and this is the amount of money he has.