The Babylonians used tablets containing multiples, squares, and reciprocals to perform the operation of multiplication. Let's look at how they would have used a table of multiples.
Suppose we want to multiply 23 × 57. A Babylonian student could have used a tablet containing the multiples of 23. Remember, Babylonains used base-60, so the table contains (1,9)_{60} for 3 × 23 instead of 69.
N 23 × N N 23 × N 1 23 6 2,18 2 46 7 2,41 3 1,9 8 3,4 4 1,32 9 3,27 5 1,55 10 3,50 N 23 × N 20 7,40 30 11,30 40 15,20 50 19,10 Multiples of 23
in base-60Now, to multiply 23 × 57, we write 57 as the sum 50 + 7 and use the distributive law to obtain
23 × 57 = 23 × (50 + 7) = 23 × 50 + 23 × 7
Using the table of muliptles at the right, we see that 23 × 50 = (19,10)_{60}
and 23 × 7 = (2,41)_{60}. Now, we simply add these two base-60 numbers to get the result of (21,51)_{60}.Amazing, the multiplication problem becomes a problem of addition! But, this makes sense since multiplication is repeated addition. Let's look at one more example:
Example: Muliplty 23 × 49 using a table of multiples.
23 × 49 = 23 × (40 + 9) = 23 × 40 + 23 × 9 = (15,20)_{60} + (3,27)_{60} = (18,47)_{60}.
Let's take a look at how they would have used a tablet containing the squares of numbers.
N N^{2} N N^{2} N N^{2} N N^{2} N N^{2} 1 1 11 2,1 21 7,21 31 16,1 41 28,1 2 4 12 2,24 22 8,4 32 17,4 42 29,24 3 9 13 2,49 23 8,49 33 18,9 43 30,49 4 16 14 3,16 24 9,36 34 19,16 44 32,16 5 25 15 3,45 25 10,25 35 20,25 45 33,45 6 36 16 4,16 26 11,16 36 21,36 46 35,16 7 49 17 4,49 27 12,9 37 22,49 47 36,49 8 1,4 18 5,24 28 13,4 38 24,4 48 38,24 9 1,21 19 6,1 29 14,1 39 25,21 49 40,1 10 1,40 20 6,40 30 15,0 40 26,40 50 41,40 Table of Squares The table at the right shows the squares of the numbers from 1 to 50 written in base-60.
There are two formulas that can be used to find a product of two numbers using squares. The first is
x×y = [(x + y)^{2} − x^{2} − y^{2}] ÷ 2
and the second is
x×y = [(x + y)^{2} − (x − y)^{2}] ÷ 4
Both formulas involve division which the Babylonians would have done by multiplying by the reciprocal, which we will look at in a moment. If all the 'digits' in a base-60 number are divisible by 2 or 4 then so is the number itself and the result is obtained by simply dividing each digit by the divisor. Let's look at an example using the first formula.
Example: Muliplty 26 × 14 using a table of squares and the formula x×y = [(x + y)^{2} − x^{2} − y^{2}] ÷ 2.
26 × 14 = [(26 + 14)^{2} − 26^{2} − 14^{2}] ÷ 2 = [40^{2} − 26^{2} − 14^{2}] ÷ 2 = [(26,40)_{60} − (11,16)_{60} − (3,16)_{60}] ÷ 2 = (12,8)_{60} ÷ 2 = (6,4)_{60}.
Let's look at an example using the second formula.
Example: Muliplty 26 × 14 using a table of squares and the formula x×y = [(x + y)^{2} − (x − y)^{2}] ÷ 4.
26 × 14 = [(26 + 14)^{2} − (26 − 14)^{2}] ÷ 4 = [40^{2} − 12^{2}] ÷ 4 = [(26,40)_{60} − (2,24)_{60}] ÷ 4 = (24,16)_{60} ÷ 4 = (6,4)_{60}.
When using the second formula, it is important to make x the larger number so you avoid negative numbers. Remember that multiplication is commutative, x × y = y × x, so the order doesn't matter and x can always be the larger number.
The Babylonians could also have multiplied the way we do today, since they used a place value system with 60 as the base. When a product is greater than 60, you carry to the next place.
Let's consider the following:
Example: Muliplty (10, 30; 40)_{60} × (0; 1, 20) using a method similar to what we do today.
10 13 10, 30; 40 × 0; 1, 20 3 30 13 20 + 10 30 40 0 14 0 53 20
Notice that 20 × 40 is 800 which is (13,20)_{60}, so you put the 20 down and carry the 13. Next 20 × 30 is 600 plus the 13 we carried is 613 which is (10,13)_{60}, so you put the 13 down and carry the 10. Next 20 × 10 is 200 plus the 10 we carried is 210 which is (3,30)_{60}, so you put the 30 down and carry the 3 which just comes down.
Since 1 times anything is the anything, the next row is easy to compute. Next, we add the columns, remembering to carry when the sum exceeds 60. Finally, count the number of sexagesimal places, in this case there are 3, one in the multiplicand and 2 in the multiplier, and place the sexagesimal point 3 places from the right.
Therefore, (10, 30; 40)_{60} × (0; 1, 20)_{60} = (14;0,53,20)_{60}.
The last example, can be written as the division problem, (10, 30; 40)_{60} ÷ 45 = (14;0,53,20)_{60}. The Babylonians performed division by multiplying by the reciprocal. For this they used a table of reciprocals (see Babylonian Table of Reciprocals). Notice that the reciprocal of 45 is (0; 1, 20)_{60}.
© 2003 by Professor William F. Widulski | The Saga of Mathematics: A Brief History |