Ancient Egyptian multiplication

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In mathematics, ancient Egyptian multiplication (also known as Egyptian multiplication, Ethiopian multiplication, Russian multiplication, or peasant multiplication), one of two multiplication methods used by scribes, is a systematic method for multiplying two numbers that does not require the multiplication table, only the ability to multiply and divide by 2, and to add. It decomposes one of the multiplicands (preferably the smaller) into a sum of powers of two and creates a table of doublings of the second multiplicand. This method may be called mediation and duplation, where mediation means halving one number and duplation means doubling the other number. It is still used in some areas.

The second Egyptian multiplication and division technique was known from the hieratic Moscow and Rhind Mathematical Papyri written in the seventeenth century B.C. by the scribe Ahmes.

Although in ancient Egypt the concept of base 2 did not exist, the algorithm is essentially the same algorithm as long multiplication after the multiplier and multiplicand are converted to binary. The method as interpreted by conversion to binary is therefore still in wide use today as implemented by binary multiplier circuits in modern computer processors.

The decomposition[edit]

The ancient Egyptians had laid out tables of a great number of powers of two, rather than recalculating them each time. The decomposition of a number thus consists of finding the powers of two which make it up. The Egyptians knew empirically that a given power of two would only appear once in a number. For the decomposition, they proceeded methodically; they would initially find the largest power of two less than or equal to the number in question, subtract it out and repeat until nothing remained. (The Egyptians did not make use of the number zero in mathematics.)

To find the largest power of 2 keep doubling your answer starting with number 1, for example

20 = 1
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32

Example of the decomposition of the number 25:

The largest power of two less than or equal to 25 is 16: 25 − 16 = 9.
The largest power of two less than or equal to 9 is 8: 9 − 8 = 1.
The largest power of two less than or equal to 1 is 1: 1 − 1 = 0.
25 is thus the sum of: 16, 8 and 1.

The table[edit]

After the decomposition of the first multiplicand, it is necessary to construct a table of powers of two times the second multiplicand (generally the smaller) from one up to the largest power of two found during the decomposition. In the table, a line is obtained by multiplying the preceding line by two.

For example, if the largest power of two found during the decomposition is 16 (as in the case of the decomposition of 25; see the example above), and the second multiplicand is 7, the table is created as follows:

1 7
2 14
4 28
8 56
16 112

The result[edit]

The result is obtained by adding the numbers from the second column for which the corresponding power of two makes up part of the decomposition of the first multiplicand. In the example above, as 25 = 16 + 8 + 1, add the corresponding multiples of 7 to get 25 × 7 = 112 + 56 + 7 = 175.

The main advantage of this technique is that it makes use of only addition, subtraction, and multiplication by two.

Example[edit]

Here, in actual figures, is how 238 is multiplied by 13. The lines are multiplied by two, from one to the next. A check mark is placed by the powers of two in the decomposition of 238.

1 13
2 26
4 52
8 104
16 208
32 416
64 832
128 1664

238 3094

Since 238 = 2 + 4 + 8 + 32 + 64 + 128, distribution of multiplication over addition gives:

238 × 13 = (128 + 64 + 32 + 8 + 4 + 2) × 13
= 128 × 13 + 64 × 13 + 32 × 13 + 8 × 13 + 4 × 13 + 2 × 13
= 1664 + 832 + 416 + 104 + 52 + 26
= 3094

Russian peasant multiplication[edit]

In the Russian peasant method, the powers of two in the decomposition of the multiplicand are found by writing it on the left and progressively halving the left column, discarding any remainder, until the value is 1 (or −1, in which case the eventual sum is negated), while doubling the right column as before. Lines with even numbers on the left column are struck out, and the remaining numbers on the right are added together.[1]

13 238
6   (remainder discarded) 476
3 952
1   (remainder discarded) 1904
     

Lines with even numbers on the left column are struck out, and the remaining numbers on the right are added, giving the answer as 3094:

13 238
6 476
3 952
1 +1904

3094
   

The algorithm can be illustrated with the binary representation of the numbers:

1101 (13) 11101110 (238)
110 (6) 111011100 (476)
11 (3) 1110111000 (952)
1 (1) 11101110000 (1904)
       
1 1 1 0 1 1 1 0 (238)
× 1 1 0 1 (13)

1 1 1 0 1 1 1 0 (238)
0 0 0 0 0 0 0 0 0 (0)
1 1 1 0 1 1 1 0 0 0 (952)
+ 1 1 1 0 1 1 1 0 0 0 0 (1904)

1 1 0 0 0 0 0 1 0 1 1 0 (3094)

See also[edit]

References[edit]

Other sources[edit]

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