Difference between revisions of "Arithmetic Textbook by R. Aharon ben Isaac"

• The first is the number deprived of substance, speech or thought, which is unlimited, because it is a number in potentia rather than in actu, and has no end.

• The second kind, which this treatise concerns, is the number initially delimited and bounded by thought. Then, through the rational capacity, man can pronounce it or write it, as one wishes. This number is thus counted and limited. It is divided into two: even and odd, whose foundation is one, because the counted number is a sum of units.

The Positional Decimal System

The First Part: Numerical Operations

The First Section: Integers

The First Chapter on Addition

• Example: Reuven owes Yehuda 484 dinar and 389 more.

We arrange them in this diagram:

Debt   Result

4 8 4 3 8 9 8 7 3

• Sum what is in the rank of the units; they result is 13.

• Example: Reuven, Shimon and Levy borrowed money from Yehuda: Reuven borrowed two thousand and eighty dinar; Shimon [borrowed] three thousand and sixty; and Levy [borrowed] four thousand and sixy. We wish to know the total amount [they borrowed].

We arrange them in this diagram:

Reuven 2 0 8 0 Shimon 3 0 6 0 Levy 4 0 6 0 Result 9 2 0 0

• Another example: we sum 248 kors of wheat with 773; the result is one thousand and 21 kors of wheat.

16	4	4		1	1	1
20	5	4		2	2	1
24	6	4		3	3	1
28	7	4		4	4	1
32	8	4		5	5	1
36	9	4		6	6	1
				ז	ז	א
כה	ה	ה		ח	ח	א
ל	ו	ה		ט	ט	א
לה	ז	ה
40	8	5		4	2	2
45	9	5		6	3	2
				8	4	2
36	6	6		10	5	2
42	7	6		12	6	2
48	8	6		14	7	2
54	9	6		16	8	2
				18	9	2
49	7	7
56	8	7		9	3	3
63	9	7		12	4	3
				15	5	3
64	8	8		18	6	3
72	9	8		21	7	3
				24	8	3
81	9	9		27	9	3

The Second Chapter on Multiplication

• Example: we wish to know [the area of] a square that is 3 cubits long and so is its width.

We multiply 3 by 3, that is, we multiply the 3 by the other 3, 3 times; the result is 9 and this is the area of the required shape.

שטח התמונה הנרצת

1	1	1
1	1	1
1	1	1

1	1	1	1
1	1	1	1
1	1	1	1
1	1	1	1

• Example: we wish to know [the area] of an equilateral square, each of whose sides is 26 cubits long.

As saying: 26 cubits of cloth at 26 dinar for one cubit; how much will it cost?

Or, as saying: 26 weights of a thing for 26 dinar, or peraḥim, or liter; how much will it cost?

We arrange them in two lines above one line in order to separate the multiplier and the multiplied from the product, as this diagram:

2 6   2 6 1 5 6 5 2   6 7 6

We multiply 6 by 6 and 6 by 2; then 2 by 6 and 2 by 2.

Because we multiply each rank in the upper row by each rank in the bottom row:

We multiply 6 by 6; the result is 36.

We write the units, which are 6, beneath the units.

The tens that are 30 are 3 in the second [rank], but we do not write them, until we make another multiplication to add with it, in order to ease the trouble.

We say: 6 times 2 is 12. With the 3 we have, it is 15.

We write the units in the second rank and the ten in the third [rank].

You get the product of 6 by the whole bottom row.

Now, we multiply 2 by the whole row as well.

We say: 2 times 6 is 12.

We write the units corresponding to its rank.

This is the required area of the equilateral square, each side of which is 26 cubits; it is also the number of dinar for the 26 cubits of cloth.

There is no need for another example, because it is as saying 26 cubits of cloth for a total of 26 dinar; the result is 676 dinar; or as saying 26 weights of copper, or something for a total of 26 dinar, or peraḥim, or liters. You can also say that about all things you want, because the multiplication of all of them is the same and so for all the examples I will give for quadrilaterals, you can change the matter of the question as you wish.

• Example: a square shape, each of whose sides is two hundred cubits long.

We arrange two hundred by two hundred above one line, as this diagram:

2 0 0     2 0 0     0 0 0   0 0 0   4 0 0     4 0 0 0 0

It is also as saying two hundred kikkar of copper at two hundred dinar for one kikkar.

We multiply the first zero in the bottom row; it is zero.

• Example: we multiply 2 by 3:

The result is 6 and this is the area of a rectangle, like this:

Because one side is 2 and the other is 3.

• Similarly, if you multiply 3 by 4:

The result is 12 and this is the area of a rectangle like this:

1 1 1 1 1 1 1 1 1 1 1 1

Because one side is 4 and the other is 3.

• Another example: we wish to know the area [of a quadrangle], one side of which is 75 and the other is 245, like this:

7 5     2 4 5   1 2 2 5 1 7 1 5   1 8 3 7 5

We multiply one by the other according to the rule: 5 by 245; the result is 1225; and 7[0] by 245; the result is 17150. We sum them up; the result is 18375.

multiplier multiplicand

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

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

product

• The entire fourth shape has different sides: It is when the three numbers are different; like a wall.

• Example: we wish [to know] the cube of 3.

We multiply 3 by 4 that follows it; it is 12. We multiply also 12 by 2, which is the preceding number; it is 24. Add the [cube] root to it; it is 27 and this is the required. Because it is the same as 3 by 3, which is 9; and 3 by 9 is 27.

• Another example: we wish [to know] the cube of 4.

We multiply 4 by 5 that follows 4; it is 20. We multiply also 20 by 3 that precedes 4; it is 60. We add to it the [cube] root, which is 4; it is 64 and this is the cube of 4.

Permutations

2 3 4 5 6 7 8 2 6 24 120 720 5040 40320

• I say that 2 generate two permutations [lit. houses], which is [the 2] in the second row beneath the 2.

• To know the permutations of 3, we multiply 3 by the permutations of 2, which is the preceding 2: there are six permutations or six combinations. So, you find that there are six combinations from 3 letters, as you see beneath the 3.

• To know the permutations of 4, we multiply [4] by 6, which is the [number of] permutations of the preceding; they are 24 beneath the 4.

• To know the permutations of 5, we multiply 5 by the preceding 24; they are 120 permutations.

• To know the combinations of 6, we multiply 6 by 120, which is the [number of] permutations of the preceding; they are 720 permutations.

• To know the combinations of 7, we multiply 7 by 720, which is the [number of] permutations of the preceding; they are 5040 permutations.

• To know the combinations of 8, we multiply 8 by 5040, which is the [number of] permutations of the preceding; they are 40320 permutations.

And so on endlessly.

RMḤ MḤR ḤRM RḤM ḤMR MRḤ

Gelosia

For example: multiply 3 by 5; it is 15 and it is in the rubric corresponding to both: write the units, which are 5, in the upper half; and 1, which is ten, at the bottom.

The same for all.

Because, the product of 7 by 5 is 35. Write 5 in the upper half and 3, which are the tens, in the other half.

As you see in the diagram: 7 by 5, by 4, by 8, and by 6; also 2 by all and 3 by all.

Then, sum them up and extract the number along the diagonal lines:

5 are the units.

8 and 3, which is 11, are the tens. Take 1 as 100 for the ten; 1 remains, which is ten.

In the third [rank] you find 6, 2, 8, 1, 5, which are 22. With the 1 we raised, they are 23 hundred. Leave 3 in its place.

We raise 20 as 2 in the fourth rank, with 2, 5, 6, 2, 1, which are 16. With the 2 we raised they are 18. Leave 8 in its place.

We raise 1 to the fifth rank, with 4, 2, 1, 4, 1, which are 12; they are 13. Leave 3 in its place.

We raise one to the sixth rank, in which there are 1, 8, 2; they are 12. We leave 2.

The Third Chapter on Subtraction

• I say that Reuven owed Shimon 573 dinars and he paid him 352. We want to subtract what he paid from the debt.

We write the two numbers by the order of their ranks, as this diagram:

Debt Repayment Remainder

5 7 3 3 5 2 2 2 1

We subtract 2 units from 3; one remains beneath the line.

We subtract 5 from 7, which are tens; 2 tens remain.

We subtract 3 from 5, which are hundreds; 2 hundreds remain.

So, 221 dinar remain from the debt.

• Another example: Reuven owed Shimon 5047 liter and he paid him 958 liter.

Debt Repayment Remainder

5 0 4 7   9 5 8 4 0 8 9

We subtract what he paid from the debt according to the previous procedure; 4089 liter remain.

Debt Repayment Remainder

9 0 0 8 5 4   4 6

We subtract 4 from zero, we cannot. We take one from the 9 and say that it is ten in the rank of units. We subtract 4 from it; 6 remains.

We subtract 5 from 9, since we considered the zero as 9, as the ten was shifted to the rank of units; 4 remains.

We subtract 8 from the 8 that remained from the 9; nothing remains.

Hence, 46 kors of wheat are left.

If you add up the payment and the remainder, you should receive the debt, as you will see in the tests.

• Another example with zeros: the amount of the debt is two hundred thousand weights of wool and he paid 123 thousand and 345.

The order of their ranks is as this diagram:

Debt Repayment Remainder

2 0 0 0 0 0 1 2 3 3 4 5   7 6 6 5 5

We cannot subtract 5 from zero, so we take 1 from the 2, shift it over all the zeros up to the first one in the rank of units and there it becomes ten. Above all the zeros in the middle, nines remain. Alternatively, consider each zero as a 9 and above the 2 remains one.

Now, we subtract 5 from the ten that we shifted, so that 5 remains; 4 from 9, so that 5 remains; and 3 from 9, so that 6 remains.

The Fourth Chapter on Division of Integers

4 8 7 9 1 2 5     7

0 0   2 4 6 2 0   1 2     1 2

2   0 3   1 5 1 4 9 7   1 4   3 4

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

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

0   0 1 2 1 5 6   1 1 1 4 4   1

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

5 2 3 7 2 7 5   1 9

תתקנ"ו כזה

4 8 4 3 8 9 8 7 3

5 4 8   1 2 4   3 6 5 1 0 3 7

3 2     7 6 5 2 4 4 8 0

The Second Section: Fractions

Introductions to the First Chapter on the Addition of Fractions

• Example: take ten, it is a half of its two sides, which are 9 and 11; 8 and 12; 7 and 13 ; and every number up to the end of the first side.

• I call [the fractions] from half to ninth a first rank that corresponds to the rank of the units.

• [The rank of] the tenth, the eleventh and so on until the ninetieth corresponds to the rank of the tens.

• Example: half is in the first rank below the one and its denominator is in the first rank above the one; and since the half is the greatest fraction, its denominator, which is two, is the smallest number.

• The next fraction after half is third, which is from the first rank below the one, and its integer denominator is three; they are secondary sides of one [...] and so on.