Reading proofs in Chinese commentaries 437
Th e way in which Liu Hui describes this process is highly interesting for our
purpose. Here is how his text reads (my emphasis):
Suppose that, when one simplifi es the circumferences of the upper and lower circles
by 3, none of the two is exhausted, then, backtracking , one makes them communi-
cate, as a consequence they are taken respectively as upper and lower diameters.
In terms of computation, the fi rst operation for multiplying quantities
with fractions is prescribed by means of the operation expressing its inten-
tion – ‘make communicate’ – which yields, respectively, 3 a (^) i + b (^) i and 3 a (^) s + b (^) s.
However, in this context , Liu Hui states, this computation carries out a
backtracking. Th is term captures two nuances. First, it refers to the fact that
one goes in a direction opposite to the one just followed. Second, it implies
that one goes back to the starting point: 3 a (^) i + b (^) i restores C (^) i , whereas 3 a (^) s + b (^) s
restores C (^) s. Two facts allow this conclusion. On the one hand, ‘making
communicate’ turns out to be the operation inverse to the division by 3,
carried out just before – and we saw how that was displayed on the count-
ing surface. On the other hand, since the results of division are given in the
form of an integer increased by a fraction, they are exact. Th is is a key fact
for ensuring that the application of the multiplication opposite to a given
division restores the original numbers – and even restores the original
set-up of the division as column 3 in Figure 13.4 shows. 22 We meet with
the importance of this key fact here for the fi rst time. We shall stress its
relevance for our topic on several occasions below.
22 Th e fact that the divisor is 3 is important to ensure that one goes back to the numbers one
started with. If simplifi cation of the remaining fraction in the result could occur, the operation
of ‘making communicate’ would not amount to applying the inverse operation.
Divisions by 3 Multiplying integers by Multiplications, sums,
corresponding denominator,
incorporating the numerator
C (^) i > D (^) i =
a
b
i
i
3
[[[ > 3ai + bi > (3ai + bi )^2 +
(3ai + bi ) (3as + bs ) +
C (^) s D (^) s =
a
b
s
s
3
[[[ 3 as + bs (3as + bs )^2
Multiplying denominators, dividing by the result, 9, ]]] , multiplying by h ,
dividing by 3