The History of Mathematical Proof in Ancient Traditions

562 tian miao

for the hypotenuse, and squares it. Finally, by eliminating the square of the

hypotenuse and the expression put on the left side, he gets the equation.

Th e second example (problem 58) shows that Li Rui deliberately followed

the same pattern in the whole book.

Suppose there is the sum of gou and the hypotenuse (equal to) 676, the diff erence

between the sum (of gu and the hypotenuse) and gou is 560. One asks how much

the same items as in the previous problem ( gou , gu and the hypotenuse) are.

Draft : set up gou as the celestial unknown, multiplying it by itself, one gets

0

0

1

, which

makes the square of gou. Further one places the sum of the hypotenuse and gou ,

676, and subtracting gou from it, one gets the following:^676

− 1

, which is the

hypotenuse. Further, one places the diff erence between the sum and gou , 560;

adding the celestial unknown gou to it, one gets the following formula,

560

1

676

, which

makes the sum of gu and the hypotenuse. Subtracting the hypotenuse, −1 from it,

one gets −^116

2

, which makes gu ; multiplying it by itself, one gets the following

formula,

13456

464

4

− , which is the square of gu. Adding the two squares together, one

gets the following formula:

13456

464

5

− , which makes the square of the hypotenuse. (Put it

on the left .) Further, multiplying the

676

− 1 hypotenuse, −1, by itself, one gets

456976

1352

1

− ,

which makes a quantity equal to (the number put aside on the left ). Eliminating the

left (number), one gets

− 443520

888

4

; dividing all the numbers from top to bottom by 4,

one gets

110880

222

1

. Solve the equation of the second degree.

One gets 240, which is the gou. Get the gu and hypotenuse according to procedure.

Th is answers the problem. 19

In modern algebra, the above outline could be reformulated into the

following procedure:

Ta k e a , the gou , as x ,

then, a^2 = x^2

as c + a = 676

19 Li Rui 1806 : 28b–29a.