The History of Mathematics: A Brief Course

2. CHINA 137

With this procedure for reducing fractions to lowest terms, a complete and

simple theory of computation with fractions is feasible. Such a theory is given in the

Sun Zi Suan Jing, including the standard procedure for converting a mixed number

to an improper fraction and the procedures for adding, subtracting, multiplying,

and dividing fractions. Thus, the Chinese had complete control over the system of

rational numbers, including, as we shall see below, the negative rational numbers.

At an early date the Chinese dealt with roots of integers, numbers like \/355,

which we now know to be irrational; and they found mixed numbers as approxima-

tions when the integer is not a perfect square. In the case of \/355, the approxima-

tion would have been given as 18||. (The denominator is always twice the integer

part, as a result of the particular algorithm used.)

From arithmetic to algebra. Sooner or later, constantly solving problems of more

and more complexity in order to find unknown quantities leads to the systematiza-

tion of ways of imagining operations performed on a "generic" number (unknown).

When the point arises at which a name or a symbol for an unknown number is

invented, so that expressions can be written representing the result of operations

on the unknown number, we may take it that algebra has arisen. There is a kind of

twilight zone between arithmetic and algebra, in which certain problems are solved

imaginatively without using symbols for unknowns, but later are seen to be easily

solvable by the systematic methods of algebra. A good example is Problem 15 of

Chapter 3 of the Sun Zi Suan Jing, which asks how many carts and how many

people are involved, given that there are two empty carts (and all the others are

full) when people are assigned three to a cart, but nine people have to walk if only

two are placed in each cart. We would naturally make this a problem in two linear

equations in two unknowns: If ÷ is the number of people and y the number of carts,

then

However, that would be using algebra, and Sun Zi does not quite do that in this

case. His solution is as follows:

Put down 2 carts, multiply by 3 to give 6, add 9, which is the

number of persons who have to walk, to obtain 15 carts. To find

the number of persons, multiply the number of carts by 2 and add

9, which is the number of persons who have to walk.

Probably the reasoning in the first sentence here is pictorial. Imagine each cart
filled with three people. When loaded in this way, the carts would accomodate all
the "real" people in the problem, plus six "fictitious" people, since we are given that
two carts would be empty if the others each carried three people. Let us imagine
then, that six of the carts contain two real people and one fictitious person, while
the others contain three real people. Now imagine one person removed from each
cart, preferably a fictitious person if possible. The number of people removed would
obviously be equal to the number of carts. The six fictitious people would then be
removed, along with the nine real people who have to walk when there are only two
people in each cart. It follows that there must be 15 carts. Finding the number of
people is straightforward once the number of carts is known.

÷ = 3(y-2)

÷ = 2y + 9.