TCM Arithmetic 2: Arithmetic of Fractions

**By: Tao Steven Zheng (鄭濤)**

## **Chinese Fractions**

In the Chinese language, fractions are verbally expressed using the characters 分之 (*fen zhi*, literally "parts of") between the denominator (分母 *fen mu*) and the numerator (分子 *fen zi*).[1] This encapsulates the concept of dividing a whole number into a certain number of equal parts. For example, the fraction three-quarters is written 四分之三 (*si fen zhi san*, literally "four parts of three"). For mixed numbers, the integer part is written first, followed by the fractional part. For example,

3

1
7

{\displaystyle 3 \frac{1}{7}}

 is written 三、七分之一 (*san, qi fen zhi yi*, literally "three, seven parts of one"). Aside from the standard expressions for fractions, there exists special names for four specific fractions that are commonly found in ancient Chinese mathematical texts (Table 1).

**Table 1. Four specific fractions**

 少半 *shao ban* (lesser half)
 中半 *zhong ban* (middle half)
 太半 *tai ban* (greater half)

[1] Ancient Chinese mathematics texts refer the denominator as the "mother" (母 *mu*) and the numerator as the "child" (子 *zi*).

## **Computation with Fractions**

The arithmeticians of ancient China were expected to be proficient with calculating fractions. In Chapter 1 of the *Jiuzhang Suanshu*, there are seven major arithmetic operations to master (Table 2). In the following problem studies, we will examine in detail the procedures and relevant terminologies for three operations: 約分術 (*yue fen shu*, method of reducing fractions), 合分術 (*he fen shu*, method of adding fractions), and 經分術 (*jing fen shu*, method of dividing fractions).

**Table 2. Arithmetic operations on fractions**

 (7) 乘分術 *cheng fen shu* (method of multiplying fractions)

### **Problem Study 1: Reducing Fractions**

Problem 5 and Problem 6 from Chapter 1 of the *Jiuzhang Suanshu* illustrates the method for obtaining the fully reduced form of a given fraction. The prescribed method for reducing fractions to lowest terms includes a process that is equivalent to the Euclidean algorithm. This process begins by making two numbers undergo a series of successive subtractions until both numbers are equal. This result is referred to as the "equal number" (等數 *deng shu*), which is equivalent to the greatest common divisor (GCD) or highest common factor (HCF).

答曰：三分之二。

[1.06] 又有九十一分之四十九。問：約之得幾何？
答曰：十三分之七。

約分術曰：可半者半之，不可半者，副置分母子之數，以少減多，更相減損，求其等也。以等數約之。

Answer: 2/3.

[1.06] Suppose there are 49/91. Question: What is the reduced form?

Answer: 7/13.

Method for reducing fractions (約分術 *yue fen shu*): If both the numerator and denominator can be halved, halve them. If they cannot be halved, set the numbers of the denominator and numerator separately, subtract the smaller from the larger, and continue subtracting until equality, to obtain the equal number (等數 *deng shu*). Use the equal number to reduce the fraction.

**Solution for *Jiuzhang Suanshu* [1.05]**

Both the numerator 12 and the denominator 18 are even numbers, so halve them.

12
÷
2

18
÷
2

=

6
9

{\displaystyle \frac{12 \div 2}{18 \div 2} = \frac{6}{9} }

Although the numerator 6 is an even number, the denominator 9 is an odd number, so they cannot both be halved. Use the Euclidean algorithm to find the equal number (等數 *deng shu*).

9
−
6

=
3

6
−
3

=
3

3

=
3

{\displaystyle
\begin{align}
9 - 6 &= 3 \\
6 - 3 &= 3 \\
3 &= 3
\end{align}
}

Thus, the equal number (等數 *deng shu*) is 3. Divide the numerator and denominator by 3 to obtain the fully reduced fraction.

6
÷
3

9
÷
3

=

2
3

{\displaystyle \frac{6 \div 3}{9 \div 3} = \frac{2}{3} }

**Solution for *Jiuzhang Suanshu* [1.06]**

The numerator 49 and the denominator 91 cannot be halved because both are odd numbers. Use the Euclidean algorithm to find the equal number (等數 *deng shu*).

91
−
49

=
42

49
−
42

=
7

42
−
7

=
35

35
−
7

=
28

28
−
7

=
21

21
−
7

=
14

14
−
7

=
7

7

=
7

{\displaystyle
\begin{align}
91 - 49 &= 42 \\
49 - 42 &= 7 \\
42 - 7 &= 35 \\
35 - 7 &= 28 \\
28 - 7 &= 21 \\
21 - 7 &= 14 \\
14 - 7 &= 7 \\
7 &= 7
\end{align}
}

Thus, the equal number (等數 *deng shu*) is 7. Divide the numerator and denominator by 7 to obtain the fully reduced fraction.

49
÷
7

91
÷
7

=

7
13

{\displaystyle \frac{49 \div 7}{91 \div 7} = \frac{7}{13} }

### **Problem Study 2: Addition of Fractions**

The following three problems (*Jiuzhang Suanshu*, Chapter 1, Problem 7 - 9) is concerned with the addition of two or more fractions. The method prescribed in the *Jiuzhang Suanshu* describes a general algorithm for computing the sums of fractions.

(1) Given an arbitrary number of fractions

a
1

b
1

,

a
2

b
2

,

a
3

b
3

,
.
.
.
,

a
n

b
n

{\displaystyle \frac{a\_1}{b\_1}, \frac{a\_2}{b\_2}, \frac{a\_3}{b\_3}, ..., \frac{a\_n}{b\_n} }

arrange the numerators and denominators as follows:

a
1

{\displaystyle a\_1}

a
2

{\displaystyle a\_2}

a
3

{\displaystyle a\_3}

a
n

{\displaystyle a\_n}
 Denominators (母 *mu*)

b
1

{\displaystyle b\_1}

b
2

{\displaystyle b\_2}

b
3

{\displaystyle b\_3 }

b
n

{\displaystyle b\_n}

(2) Calculate the adjusted numerators for each fraction by using a cross-multiplication procedure called *hu cheng* (互乘). This process is calculated by multiplying each numerator by the denominators of the other fractions. Chinese mathematicians would later call this operation "homogenizing" (齊 *qi*).

A
1

{\displaystyle A\_1}

a
1

×

b
2

×

b
3

×
.
.
.
×

b
n

{\displaystyle a\_1 \times b\_2 \times b\_3 \times ... \times b\_n }

A
2

{\displaystyle A\_2}

a
2

×

b
1

×

b
3

×
.
.
.
×

b
n

{\displaystyle a\_2 \times b\_1 \times b\_3 \times ... \times b\_n }

A
3

{\displaystyle A\_3}

a
3

×

b
1

×

b
2

×
.
.
.
×

b
n

{\displaystyle a\_3 \times b\_1 \times b\_2 \times ... \times b\_n }

⋮

{\displaystyle \vdots}

⋮

{\displaystyle \vdots}

A
n

{\displaystyle A\_n}

a
n

×

b
1

×

b
2

×
.
.
.
×

b

n
−
1

{\displaystyle a\_n \times b\_1 \times b\_2 \times ... \times {b}\_{n-1} }

(3) Add all the *hu cheng* (互乘) products together to obtain the dividend

A

{\displaystyle A}

.

# A

A
1

A
2

A
3

.
.
.

A
n

{\displaystyle A = A\_1 + A\_2 + A\_3 + ... + A\_n }

(4) Calculate the divisor

B

{\displaystyle B}

 by multiplying all the given denominators:

# B

b
1

×

b
2

×

b
3

×
.
.
.
×

b
n

{\displaystyle B = b\_1 \times b\_2 \times b\_3 \times ... \times b\_n }

Chinese mathematicians would later call this operation "equalizing" (同 *tong*). Together with "homogenizing " (齊 *qi*) , both operations forms the "homogenizing and equalization method" (齊同術 *qi tong shu*).

(5) Divide the dividend (實 *shi*) by the divisor (法 *fa*):

A
B

{\displaystyle \frac{A}{B}}

(6) Lastly, reduce the fraction to lowest terms if needed:

A
B

→

a
b

{\displaystyle \frac{A}{B} \rightarrow \frac{a}{b} }

where

g
c
d
(
a
,
b
# )
1

{\displaystyle gcd(a,b) = 1}

. If

a
>
b

{\displaystyle a > b}

, express the fraction as a mixed number

a
b

=
q

r
b

{\displaystyle \frac{a}{b} = q + \frac{r}{b} }

, where

# a
b
q
r

{\displaystyle a=bq+r}

 and

0
<
r
<
b

{\displaystyle 0 < r < b }

.

答曰：十五分之十一。

[1.08] 又有三分之二，七分之四，九分之五。問：合之得幾何？
答曰：得一、六十三分之五十。

[1.09] 又有二分之一，三分之二，四分之三，五分之四。問：合之得幾何？
答曰：得二、六十分之四十三。

合分術曰：母互乘子，并以為實，母相乘為法，實如法而一。不滿法者，以法命之。其母同者，直相從之。

1
3

{\displaystyle \frac{1}{3}}

 and

2
5

{\displaystyle \frac{2}{5}}

 . Question: What is the sum?

Answer:

11
15

{\displaystyle \frac{11}{15} }

.

[1.08] Again, suppose there are

2
3

{\displaystyle \frac{2}{3}}

,

4
7

{\displaystyle \frac{4}{7} }

,

5
9

{\displaystyle \frac{5}{9}}

. Question: What is the sum?

Answer: The result is

1

50
63

{\displaystyle 1\frac{50}{63} }

.

[1.09] Again, suppose there are

1
2

{\displaystyle \frac{1}{2}}

,

2
3

{\displaystyle \frac{2}{3}}

,

3
4

{\displaystyle \frac{3}{4}}

,

4
5

{\displaystyle \frac{4}{5}}

. Question: What is the sum?

Answer: The result is

2

43
60

{\displaystyle 2\frac{43}{60} }

 .

Method for adding fractions (合分術 *he fen shu*): Cross-multiply (互乘 *hu cheng*) the numerators and denominators, and combine them as the dividend (實 *shi*). Mutually multiply all of the denominators together as the divisor (法 *fa*). Divide the dividend by the divisor; if it does not divide completely, then use the divisor to name (命 *ming*) the fraction. If their denominators are the same, then directly add them together.

**Solution for *Jiuzhang Suanshu* [1.07]**

1

{\displaystyle 1}

2

{\displaystyle 2}
 Denominators (母 *mu*)

3

{\displaystyle 3}

5

{\displaystyle 5}

(1) Calculate the dividend (實 *shi*).

A

=
1
×
5
2
×
3

=
5
6

=
11

{\displaystyle
\begin{align}
A &= 1 \times 5 + 2 \times 3 \\
&= 5 + 6 \\
&= 11
\end{align}}

Divisor (法 *fa*):

# B
3
×
5

{\displaystyle B = 3 \times 5 }

Divide the dividend (實 *shi*) by the divisor (法 *fa*):

A
B

=

11
15

{\displaystyle \frac{A}{B} = \frac{11}{15}}

**Solution for *Jiuzhang Suanshu* [1.08]**

2

{\displaystyle 2}

4

{\displaystyle 4}

5

{\displaystyle 5}
 Denominators (母 *mu*)

3

{\displaystyle 3}

7

{\displaystyle 7}

9

{\displaystyle 9}

Dividend (實 *shi*):

A

=
2
×
7
×
9
4
×
3
×
9
5
×
3
×
7

=
126
108
105

=
339

{\displaystyle
\begin{align}
A &= 2 \times 7 \times 9 + 4 \times 3 \times 9 + 5 \times 3 \times 7 \\
&= 126 + 108 + 105 \\
&= 339
\end{align}}

Divisor (法 *fa*):

# B
3
×
7
×
# 9
189

{\displaystyle B = 3 \times 7 \times 9 = 189 }

Divide the dividend (實 *shi*) by the divisor (法 *fa*):

A
B

=

339
189

=

113
63

=
1

50
63

{\displaystyle \frac{A}{B} = \frac{339}{189} = \frac{113}{63} = 1\frac{50}{63} }

**Solution for *Jiuzhang Suanshu* [1.09]**

1

{\displaystyle 1}

2

{\displaystyle 2}

3

{\displaystyle 3}

4

{\displaystyle 4}
 Denominators (母 *mu*)

2

{\displaystyle 2}

3

{\displaystyle 3}

4

{\displaystyle 4}

5

{\displaystyle 5}

Dividend (實 *shi*):

A

=
1
×
3
×
4
×
5
2
×
2
×
4
×
5
3
×
2
×
3
×
5
4
×
2
×
3
×
4

=
60
80
90
96

=
326

{\displaystyle
\begin{align}
A &= 1 \times 3 \times 4 \times 5 + 2 \times 2 \times 4 \times 5 + 3 \times 2 \times 3 \times 5 + 4 \times 2 \times 3 \times 4 \\
&= 60 + 80 + 90 + 96 \\
&= 326
\end{align}}

Divisor (法 *fa*):

# B
2
×
3
×
4
×
# 5
120

{\displaystyle B = 2 \times 3 \times 4 \times 5 = 120 }

Divide the dividend (實 *shi*) by the divisor (法 *fa*):

A
B

=

326
120

=

163
60

=
2

43
60

{\displaystyle \frac{A}{B} = \frac{326}{120} = \frac{163}{60} = 2\frac{43}{60} }

### **Problem Study 3: Division of Fractions**

Problem 17 and Problem 18 from Chapter 1 of the *Jiuzhang Suanshu* focuses on the division of fractions. Each problem is about splitting a given quantity of money in units of *qian* (錢, literally "cash" or "coin") among a given number of people. In Problem 18, the money is divided among

3

1
3

{\displaystyle 3\frac{1}{3} }

people; hence, the amount of money and number of people should be treated as abstract quantities.

答曰：人得一錢、二十一分錢之四。

[1.18] 又有三人，三分人之一，分六錢三分錢之一，四分錢之三。問：人得幾何？
答曰：人得二錢、八分錢之一。

經分術曰：以人數為法，錢數為實，實如法而一。有分者通之，重有分者同而通之。

8

1
3

{\displaystyle 8\frac{1}{3} }

 *qian*. Question: How much should each person receive?

Answer: Each person receives

1

4
21

{\displaystyle 1 \frac{4}{21} }

 *qian*.

[1.18] Again, suppose there are

3

1
3

{\displaystyle 3\frac{1}{3} }

 people dividing

6

1
3

{\displaystyle 6\frac{1}{3} }

 *qian* and

3
4

{\displaystyle \frac{3}{4}}

 *qian*. Question: How much should each person receive?

Answer: Each person receives

2

1
8

{\displaystyle 2 \frac{1}{8} }

 *qian*.

Method for dividing fractions (經分術 *jing fen shu*): Use the number of people as the divisor, and the amount of *qian* as the dividend; divide the dividend by the divisor. If there are mixed numbers, interconnect (通 *tong*) them. If both of the rates are mixed numbers, then interconnect and equalize them.

Prior to the Han dynasty, the mathematicians of ancient China already understood how to multiply and divide fractions. They were aware that dividing two fractions

a
b

{\displaystyle \frac{a}{b}}

 and

c
d

{\displaystyle \frac{c}{d}}

 is equivalent to taking the first fraction and multiplying it with the reciprocal of the second fraction.

a
b

÷

c
d

=

a
b

×

d
c

{\displaystyle \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} }

In the *Jiuzhang Suanshu*, the term "interconnect" (通 *tong*) refers to the procedure of converting a mixed number into an improper fraction.

a

b
c

→

a
c
b

c

{\displaystyle a \frac{b}{c} \rightarrow \frac{ac+b}{c}}

**Solution for *Jiuzhang Suanshu* [1.17]**

8

1
3

÷
7

=

25
3

×

1
7

=

25
21

=
1

4
21

{\displaystyle
\begin{align}
8 \frac{1}{3} \div 7 &= \frac{25}{3} \times \frac{1}{7} \\
&= \frac{25}{21} \\
&= 1 \frac{4}{21}
\end{align}}

**Solution for *Jiuzhang Suanshu* [1.18]**

(

6

1
3

3
4

)

÷
3

1
3

=
7

1
12

÷
3

1
3

=

85
12

÷

10
3

=

85
12

×

3
10

=

17
4

×

1
2

=

17
8

=
2

1
8

{\displaystyle
\begin{align}
\left(6 \frac{1}{3} + \frac{3}{4} \right) \div 3 \frac{1}{3} &= 7 \frac{1}{12} \div 3 \frac{1}{3} \\
&= \frac{85}{12} \div \frac{10}{3} \\
&= \frac{85}{12} \times \frac{3}{10} \\
&= \frac{17}{4} \times \frac{1}{2} \\
&= \frac{17}{8} \\
&= 2 \frac{1}{8}
\end{align}}

## **References**

[1] 钱宝琮主编.《中国数学史》，北京：科学出版社，1964年。

[2] 纪志刚《孙子算经、张邱建算经、夏侯阳算经导读》， 武汉：湖北教育出版社，1999年。

[3] 郭书春《九章算术译注》，上海：上海古籍出版社，2009年。

[4] 李兆华《中国数学史基础》，天津：天津教育出版社，2010年。

[5] Martzloff, Jean-Claude. A History of Chinese Mathematics (English Translation). Spinger-Verlag. 1997, 2006.

[6] Yong, Lam Lay, & Ang, Tian Se. *Fleeting Footsteps: Tracing the Conception of Arithmetic and Algebra in Ancient China, Revised Edition*. World Scientific Publishing Company. 2004

[7] Wilkinson, Endymion. *Chinese History: A New Manual, 4th Edition*. Harvard University Press. 2015.