142 6. CALCULATIONTherefore, the two numbers are 8.5 + 4.5 = 13 and 8.5-4.5 = 4. The two techniques
just illustrated occur constantly in the cuneiform texts, and seem to be procedures
familiar to everyone, requiring no explanation.5. The ancient Greeks
It is fairly obvious how to do addition and subtraction within the Greek system
of numbering. Doing multiplication in modern notation, as you know, involves
memorizing all products up to 9 • 9, and learning how to keep columns straight,
plus carrying digits where necessary. With our place-value notation, we have little
difficulty multiplying, say 23-42. For the ancient Greeks the corresponding problem
of multiplying ê'ç' by ì' â' was more complicated. It would be necessary to find
four products: ê' • ì', ê' • â', ç' • ì', and 7' • â' • The first of three of these require
the one doing the calculation to keep in mind that ê' is 10 times â' and ì' is 10
times ä'.
These operations are easier for us, since we use the same nine digits in different
contexts, keeping track of the numbers they represent by keeping the columns
straight while multiplying. They were more difficult for the ancient Greeks, since
going from 30 to 3 was not merely a matter of ignoring a zero; it involved a shift
forward by 10 letters in the alphabet, from ë' to 7'. In addition to the carrying that
we must do, the Greek calculator had to commit to memory 20 such alphabet shifts
(10 by 10 letters, and 10 shifts by 20 letters), and, while computing, remember how
many such shifts of each kind were performed, so as to know how many factors of
10 were being "stored" during the calculation.
The procedure is explained in the Synagoge of the third-century mathematician
Pappus of Alexandria. The surviving portion of this work begins in the middle
of Book 2, explaining how to multiply quickly numbers that are multiples of 10.
Pappus illustrates the procedure by the following example, in which number names
are translated into English but number symbols are transcribed directly from the
original. (See Fig. 3 in Chapter 5 for the numerical values represented by these
Greek letters.)
Let the numbers be v', v', v', ì', ì', and ë'. Then the basic
numbers will be å', å', å', ä', ä', and 7'· Their product will be
éò'. Since there are ò' tens, and since ò' divided by four leaves a
remainder of two, the product [of the six reduced numbers] will
contain altogether one hundred myriads... and these p' myriads
multiplied by the <ò' units will make î' twofold [that is, "square"]
myriads.When we translate the problem into our notation, it becomes trivial. We are
trying to find the product 50 · 50 · 50 · 40 • 40 · 30. We have no trouble factoring out
the six zeros and rewriting the problem as5-5-5-4-4-31,000,000 = 6000 · 10^6 =
60 · (10,000)^2. Converting from 40 to 4 is considerably easier than converting from
ì' to ä'. Pappus divided the number of tens by 4, since he was counting in myriads
(10^4 ), and he expressed the answer as "myriadon î' diplon," that is, "of myriads
60 twofold," or, in better English, "60 twofold myriads." To describe what we now
call the square of a number, the ancient Greeks had to extend the normal meaning
of the word diploos (double). A nonmathematical reader of Pappus' Greek might