G7 Tutorial 01: Calculations with Rational Numbers

Key Concepts and Methods

Operations with rational numbers are rich in content, flexible in methods, and strong in techniques. They are the most basic operations in junior high school algebra. During the operation process, skillfully using operation rules (such as commutative and associative laws for addition, commutative and associative laws for multiplication, and the distributive law) and other operation methods and techniques can make operations simple and convenient.

Examples

Example 1. Calculate: \[29 \frac{3}{5} - 1 \frac{1}{3} - 15 \frac{1}{4} + 3 \frac{2}{3} - 2 \frac{1}{3} - 14 \frac{2}{5} + 0.25\]

Solution: Original expression: \[ \begin{aligned} &= \left(-1 \frac{1}{3} + 3 \frac{2}{3} - 2 \frac{1}{3}\right) + \left(29 \frac{3}{5} - 14 \frac{2}{5}\right) + \left(-15 \frac{1}{4} + \frac{1}{4}\right) \\ &= 0 + (29 - 14 - 15) + \left(\frac{3}{5} - \frac{2}{5}\right) + \left(-\frac{1}{4} + \frac{1}{4}\right) \\ &= 0 + \frac{1}{5} + 0 \\ &= \frac{1}{5} \end{aligned} \]

Explanation: When performing addition and subtraction with rational numbers, first observe if there are numbers that sum to 0. If there are, combine them first. Then, numbers with the same denominator can be added together. If it’s a mixed fraction, its integer part and fractional part can be combined separately. If there are both decimals and fractions, usually convert the decimals to fractions. When rational numbers are exchanged or combined, they must carry their own sign (“+” or “-” sign) when exchanged or combined.

Example 2. Calculate: \(-2.5 \div 0.75 \times \left(-\frac{1}{5}\right) \times \left(-1 \frac{3}{4}\right) \div (-1.4) \times \left(-\frac{3}{5}\right) \times \frac{2}{3}\)

Solution: Original expression: \[ \begin{aligned} &= -\frac{5}{2} \div \frac{3}{4} \times \left(-\frac{1}{5}\right) \times \left(-\frac{7}{4}\right) \div \left(-\frac{7}{5}\right) \times \left(-\frac{3}{5}\right) \times \frac{2}{3} \\ &= -\left(\frac{5}{2} \times \frac{4}{3} \times \frac{1}{5} \times \frac{7}{4} \times \frac{5}{7} \times \frac{3}{5} \times \frac{2}{3}\right) \\ &= -\frac{1}{3} \end{aligned} \]

Explanation: When performing multiplication and division operations with rational numbers, pay attention to determining the sign of the result. If there is an odd number of negative factors, the result is negative; if there is an even number of negative factors, the result is positive. Usually, decimals are converted to fractions, mixed numbers are converted to improper fractions, and division is converted to multiplication. Cancel common factors if possible. In addition, memorize some commonly used data, such as \(0.125 = \frac{1}{8}\), \(0.25 = \frac{1}{4}\), \(0.375 = \frac{3}{8}\), \(0.75 = \frac{3}{4}\), etc.

Example 3. Calculate: \(0.7 \times 1 \frac{2}{11} - 6.6 \times \frac{3}{7} - 2.2 \div \frac{7}{3} + 0.7 \times \frac{9}{11} + 3.3 \div \frac{7}{8}\)

Solution: Original expression: \[ \begin{aligned} &= \left(0.7 \times 1 \frac{2}{11} + 0.7 \times \frac{9}{11}\right) + \left(-6.6 \times \frac{3}{7} - 2.2 \times \frac{3}{7} + 3.3 \times \frac{8}{7}\right) \\ &= 0.7 \times \left(1 \frac{2}{11} + \frac{9}{11}\right) - 6.6 \times \left(\frac{3}{7} + \frac{1}{7} - \frac{4}{7}\right) \\ &= 0.7 \times 2 - 6.6 \times 0 \\ &= 1.4 \end{aligned} \]

Explanation: If there are common factors, generally factor them out first. Sometimes factors can be used by finding internal relationships (e.g., \(6.6, 3.3, 2.2\) all share \(1.1\) or \(2.2\)). If the common factor is negative, pay special attention to the signs of all terms changing after factoring.

Example 4. Calculate: \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}\)

Solution: Original expression: \[ \begin{aligned} &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \left(\frac{1}{64} + \frac{1}{64}\right) - \frac{1}{64} \\ &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \left(\frac{1}{32} + \frac{1}{32}\right) - \frac{1}{64} \\ &= \dots \\ &= \left(\frac{1}{2} + \frac{1}{2}\right) - \frac{1}{64} \\ &= 1 - \frac{1}{64} \\ &= \frac{63}{64} \end{aligned} \]

Explanation: Each term is half of the previous term. Adding an auxiliary number \(\frac{1}{64}\) allows the sequence to collapse, provided we subtract it at the end.

Method 2: Let the original expression be \(S\). Multiplying both sides by \(64\): \[64S = 32 + 16 + 8 + 4 + 2 + 1 = 63\] \[S = \frac{63}{64}\]

Method 4: Let the original expression be \(S\). Multiplying both sides by \(\frac{1}{2}\): \[\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128}\] Then, \(S - \frac{1}{2}S\): \[ \begin{aligned} \frac{1}{2}S &= \frac{1}{2} - \frac{1}{128} = \frac{63}{128} \\ S &= \frac{63}{64} \end{aligned} \]

Example 5. Calculate: 1. \(1 + 2 + 3 + 4 + \cdots + 1997 + 1998 + 1999\) 2. \(1 - 2 + 3 - 4 + \cdots + 1997 - 1998 + 1999\)

Solution: 1. Let \(S = 1 + 2 + \dots + 1999\). Also \(S = 1999 + 1998 + \dots + 1\). Adding them: \[2S = (1+1999) \times 1999 = 2000 \times 1999\] \[S = \frac{2000 \times 1999}{2} = 1999000\]

Explanation: This is an Arithmetic Progression. \[\text{Sum} = \frac{(\text{First term} + \text{Last term}) \times \text{Number of terms}}{2}\] \[\text{Number of terms} = \frac{\text{Last term} - \text{First term}}{\text{Common difference}} + 1\]

2. Original expression \(= 1 + (-2+3) + (-4+5) + \cdots + (-1998+1999)\) \[= 1 + \underbrace{1 + 1 + \cdots + 1}_{1000 \text{ ones}} = 1000\]

Example 6. Calculate: \(\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{1999 \times 2000}\)

Solution: Original expression: \[ \begin{aligned} &= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{1999} - \frac{1}{2000}\right) \\ &= 1 - \frac{1}{2000} = \frac{1999}{2000} \end{aligned} \]

Explanation: This is the partial fraction decomposition method. \[\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\] Other forms: 1. \(\frac{1}{n(n+d)} = \frac{1}{d} \left(\frac{1}{n} - \frac{1}{n+d}\right)\) 2. \(\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right]\)

Example 7. \(1999\) plus \(\frac{1}{2}\) of itself gives a number. Then plus \(\frac{1}{3}\) of the obtained number gives another number. This continues until it is added to \(\frac{1}{1999}\) of the previously obtained number. What is the final number?

Solution: The final number is: \[ \begin{aligned} & 1999 \times \left(1 + \frac{1}{2}\right) \times \left(1 + \frac{1}{3}\right) \times \cdots \times \left(1 + \frac{1}{1999}\right) \\ &= 1999 \times \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{2000}{1999} \\ &= 1999 \times \frac{2000}{2} = 1999000 \end{aligned} \]

Exercises

Quiz

1. \(3 \frac{1}{6} - 4 \frac{1}{7} + 5 \frac{1}{8} - 6 \frac{1}{6} + 7 \frac{1}{7} - 8 \frac{1}{8}\)
2. \(2.2 \times (-2.1) + 1.21 \times 4.2 - 2.1 \times 0.22\)
3. \(1 \frac{1}{2} + 3 \frac{1}{4} + 5 \frac{1}{8} + \cdots + 17 \frac{1}{512}\)
4. \(1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + \cdots + 1997 + 1998 - 1999 - 2000\)
5. \(1 + 1 \frac{1}{4} + 1 \frac{1}{2} + 1 \frac{3}{4} + 2 + \cdots + 20\)
6. \(\frac{1}{1 \times 6} + \frac{1}{6 \times 11} + \frac{1}{11 \times 16} + \cdots + \frac{1}{51 \times 56}\)
7. \(\frac{1}{2} + \frac{1}{2+4} + \frac{1}{2+4+6} + \cdots + \frac{1}{2+4+\cdots+200}\)
8. \((-1)^1 + (-1)^2 + \cdots + (-1)^{1999}\)
9. \(1^2 - 2^2 + 3^2 - 4^2 + \cdots + 1997^2 - 1998^2 + 1999^2\)
10. \(1 + \frac{1}{3} + \frac{1}{3^2} + \cdots + \frac{1}{3^{16}}\)
11. Average Score: There are 20 students with scores: 86, 91, 93, 87, 88, 94, 87, 93, 95, 89, 88, 92, 96, 85, 83, 93, 97, 81, 89, 94. Find the average.
12. Remainder Problem: \(1999\) minus \(\frac{1}{2}\) of itself, then minus \(\frac{1}{3}\) of the remainder, \(\dots\), until \(\frac{1}{1999}\) of the remainder is subtracted. What is the final number remaining?