Classwise Concept with Examples
6th	7th	8th	9th	10th	11th	12th

Class 8th Chapters
1. Rational Numbers	2. Linear Equations in One Variable	3. Understanding Quadrilaterals
4. Practical Geometry	5. Data Handling	6. Squares and Square Roots
7. Cubes and Cube Roots	8. Comparing Quantities	9. Algebraic Expressions and Identities
10. Visualising Solid Shapes	11. Mensuration	12. Exponents and Powers
13. Direct and Inverse Proportions	14. Factorisation	15. Introduction to Graphs
16. Playing with Numbers

Content On This Page
Rational Numbers	Addition of Rational Numbers	Subtraction of Rational Numbers
Multiplication of Rational Numbers	Division of Rational Numbers	Representation of Rational Numbers on the Number Line
Rational Numbers Between Two Rational Numbers	Word Problems Related to Rational Numbers

Chapter 1 Rational Numbers (Concepts)

Welcome back to the exploration of our number system! This chapter significantly revisits and deepens our understanding of Rational Numbers, a crucial set of numbers first formally introduced in Class 7. We will reinforce the fundamental definition and then delve more formally into the intricate algebraic structure that governs how these numbers behave under various mathematical operations. Think of this as moving beyond simply knowing *what* rational numbers are, to understanding *how* they consistently interact according to specific mathematical laws.

Let's begin by firmly re-establishing the definition: a rational number is any number that can be expressed precisely in the form $\frac{p}{q}$, where $p$ represents the numerator and $q$ represents the denominator. Both $p$ and $q$ must be integers (..., -2, -1, 0, 1, 2, ...), and there's a vital condition: the denominator $q$ cannot be zero ($q \neq 0$). This definition elegantly encompasses integers ($n = \frac{n}{1}$), terminating decimals (like $0.25 = \frac{1}{4}$), and non-terminating repeating decimals (like $0.\overline{6} = \frac{2}{3}$).

A primary focus of this chapter is the formal examination of the properties of rational numbers under the fundamental arithmetic operations: addition, subtraction, multiplication, and division. These properties ensure predictability and consistency when performing calculations:

Closure Property: Rational numbers are 'closed' under addition, subtraction, and multiplication. This means that if you add, subtract, or multiply any two rational numbers, the result will always be another rational number. They are also closed under division, with the critical exception that division by zero ($\frac{p}{0}$) is undefined.
Commutativity: The order in which you add or multiply two rational numbers does not affect the result. For any rational numbers $a$ and $b$:
- $a + b = b + a$ (Commutativity of Addition)
- $a \times b = b \times a$ (Commutativity of Multiplication)
Note: Subtraction and division are generally not commutative for rational numbers.
Associativity: When adding or multiplying three or more rational numbers, the way you group them using parentheses does not change the outcome. For any rational numbers $a$, $b$, and $c$:
- $(a + b) + c = a + (b + c)$ (Associativity of Addition)
- $(a \times b) \times c = a \times (b \times c)$ (Associativity of Multiplication)
Note: Subtraction and division are generally not associative.

We also formally recognize the roles of special numbers within the rational system:

The number 0 is the Additive Identity because adding $0$ to any rational number $a$ leaves it unchanged: $a + 0 = 0 + a = a$.
The number 1 is the Multiplicative Identity because multiplying any rational number $a$ by $1$ leaves it unchanged: $a \times 1 = 1 \times a = a$.

The concept of inverses is crucial for solving equations and understanding operations:

Every rational number $\frac{p}{q}$ has an Additive Inverse (or negative), denoted as $-\frac{p}{q}$, such that their sum is the additive identity: $\frac{p}{q} + (-\frac{p}{q}) = 0$.
Every non-zero rational number $\frac{p}{q}$ (where $p \neq 0$) has a Multiplicative Inverse (or reciprocal), denoted as $\frac{q}{p}$, such that their product is the multiplicative identity: $(\frac{p}{q}) \times (\frac{q}{p}) = 1$.

Furthermore, the Distributive Property of Multiplication over Addition provides a vital link between these two operations: $a \times (b + c) = (a \times b) + (a \times c)$. This property is exceptionally useful for simplifying complex expressions involving rational numbers.

Building on previous skills, we will practice representing rational numbers on the number line with increased precision. A significant reinforcement is the exploration of the density property of rational numbers: between any two distinct rational numbers, no matter how close, there exist infinitely many other rational numbers. We will employ methods like finding the mean (average) of two numbers, say $x$ and $y$, as $\frac{x+y}{2}$, or by generating equivalent fractions with larger common denominators to systematically find rational numbers lying between two given ones. This chapter solidifies your grasp of the rational number system as a coherent algebraic structure, preparing you for more advanced mathematical concepts.

Rational Numbers

Welcome to the world of Rational Numbers! Before we define what a rational number is, let's quickly revisit the number systems you are already familiar with.

Revision of Number Systems

We have previously learned about different collections of numbers:

1. Natural Numbers:

These are the numbers we use for counting everyday objects. They start from 1 and go on infinitely.

Set of Natural Numbers, denoted by $\mathbb{N}$, is $\{1, 2, 3, 4, ...\}$.

2. Whole Numbers:

If we include zero (0) along with the natural numbers, we get the collection of whole numbers.

Set of Whole Numbers, denoted by $\mathbb{W}$, is $\{0, 1, 2, 3, 4, ...\}$.

Every natural number is a whole number, but not every whole number is a natural number (since 0 is a whole number but not a natural number).

3. Integers:

When we include the negative counterparts of all natural numbers (..., -3, -2, -1) along with the whole numbers (0, 1, 2, 3, ...), we get the collection of integers.

Set of Integers, denoted by $\mathbb{Z}$ (from the German word 'Zahlen' meaning 'numbers'), is $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

Every natural number and every whole number is an integer.

What is a Rational Number?

Rational numbers are an extension of integers and include all numbers that can be expressed as a fraction.

A number is called a rational number if it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, and the denominator $q$ is not equal to zero ($q \neq 0$).

The word 'rational' comes from the word 'ratio'. A rational number is essentially a number that can be expressed as a ratio of two integers.

Let's see if various types of numbers fit this definition:

Positive Fractions: $\frac{3}{4}$. Here, $p=3$ (an integer) and $q=4$ (an integer, not zero). Yes, $\frac{3}{4}$ is a rational number.
Negative Fractions: $-\frac{5}{7}$. This can be written as $\frac{-5}{7}$. Here, $p=-5$ (an integer) and $q=7$ (an integer, not zero). Yes, $-\frac{5}{7}$ is a rational number.
Integers: Consider the integer 6. Can we write it as $\frac{p}{q}$? Yes, we can write $6 = \frac{6}{1}$. Here, $p=6$ and $q=1$. Both are integers, and $q=1 \neq 0$. Similarly, $-2 = \frac{-2}{1}$, $0 = \frac{0}{1}$ (or $\frac{0}{5}$, $\frac{0}{-3}$ etc., as long as the denominator is a non-zero integer). Thus, every integer is a rational number. This means Natural numbers and Whole numbers are also rational numbers, as they are subsets of integers.
Terminating Decimals: Consider $0.5$. This can be written as $\frac{5}{10}$, which simplifies to $\frac{1}{2}$. Here, $p=1$ and $q=2$. Both are integers, $q \neq 0$. So, $0.5$ is a rational number. Similarly, $2.75 = \frac{275}{100} = \frac{11}{4}$. All terminating decimals can be written as fractions $\frac{p}{q}$.
Non-terminating Repeating Decimals: Consider $0.\overline{3} = 0.333...$. As you might have learned or will learn, this can be expressed as the fraction $\frac{1}{3}$. Here, $p=1$ and $q=3$. Both are integers, $q \neq 0$. So, $0.\overline{3}$ is a rational number. Similarly, $0.\overline{14} = \frac{14}{99}$, $1.\overline{6} = \frac{16-1}{9} = \frac{15}{9} = \frac{5}{3}$. All non-terminating repeating decimals can be expressed in the form $\frac{p}{q}$.

Numbers that cannot be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ are called irrational numbers. Examples include $\sqrt{2}, \sqrt{3}, \pi$. You will study these in detail in higher classes.

Equivalent Rational Numbers

Just like equivalent fractions, different fractional forms can represent the same rational number. For example, $\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \frac{10}{20}, \frac{-5}{-10}$ all represent the same rational number. These are called equivalent rational numbers.

To get an equivalent rational number, you can multiply or divide both the numerator and the denominator by the same non-zero integer:

$\frac{p}{q} = \frac{p \times m}{q \times m}$, where $m$ is a non-zero integer.

... (i)

$\frac{p}{q} = \frac{p \div n}{q \div n}$, where $n$ is a common divisor of $p$ and $q$, and $n \neq 0$.

... (ii)

Example: $\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$. So, $\frac{2}{3}$ and $\frac{4}{6}$ are equivalent rational numbers.

Standard Form of a Rational Number

A rational number $\frac{p}{q}$ is said to be in its standard form (or simplest form or lowest terms) if:

The denominator $q$ is a positive integer.
The numerator $p$ and the denominator $q$ have no common factor other than 1. This means their Highest Common Factor (HCF) is 1, i.e., $p$ and $q$ are coprime.

Every rational number can be reduced to its standard form by dividing both the numerator and the denominator by their HCF. If the denominator is negative, first make it positive by multiplying both numerator and denominator by -1.

$\frac{p}{-q} = \frac{p \times (-1)}{-q \times (-1)} = \frac{-p}{q}$

Example 1. Reduce the rational number $\frac{36}{48}$ to its standard form.

Answer:

Given rational number: $\frac{36}{48}$.

Step 1: Find the Highest Common Factor (HCF) of the numerator (36) and the denominator (48).

Prime factors of 36: $2 \times 2 \times 3 \times 3$

Prime factors of 48: $2 \times 2 \times 2 \times 2 \times 3$

The common prime factors are $2, 2, 3$.

HCF of 36 and 48 is $2 \times 2 \times 3 = 12$.

Step 2: Divide both the numerator and the denominator by their HCF.

$\frac{36}{48} = \frac{36 \div 12}{48 \div 12}$

$= \frac{3}{4}$

The resulting rational number $\frac{3}{4}$ has a positive denominator (4), and the numerator (3) and denominator (4) have no common factors other than 1 (HCF is 1). So, $\frac{3}{4}$ is the standard form.

Alternatively, you can simplify by dividing by common factors iteratively:

$\frac{36}{48} = \frac{\cancel{36}^{\normalsize 18}}{\cancel{48}_{\normalsize 24}}$

$= \frac{18}{24} = \frac{\cancel{18}^{\normalsize 9}}{\cancel{24}_{\normalsize 12}}$

$= \frac{9}{12} = \frac{\cancel{9}^{\normalsize 3}}{\cancel{12}_{\normalsize 4}}$

$= \frac{3}{4}$

Example 2. Reduce the rational number $\frac{-25}{45}$ to its standard form.

Answer:

Given rational number: $\frac{-25}{45}$.

Step 1: Check if the denominator is positive. Yes, 45 is positive.

Step 2: Find the HCF of the absolute values of the numerator (25) and the denominator (45).

Prime factors of 25: $5 \times 5$

Prime factors of 45: $3 \times 3 \times 5$}

The common prime factor is 5.

HCF of 25 and 45 is 5.

Step 3: Divide both the numerator and the denominator by their HCF.

$\frac{-25}{45} = \frac{-25 \div 5}{45 \div 5}$

$= \frac{-5}{9}$

The resulting rational number $\frac{-5}{9}$ has a positive denominator (9), and the numerator (-5) and denominator (9) have no common factors other than 1 (HCF is 1). So, $\frac{-5}{9}$ is the standard form.

Addition of Rational Numbers

Now that we understand what rational numbers are, let's learn how to perform the basic arithmetic operation of addition with them. The method for adding rational numbers is similar to adding fractions, which you have already learned.

Case 1: Adding Rational Numbers with the Same Denominators

When two or more rational numbers have the same denominator (called like rational numbers), the addition is straightforward. We simply add their numerators and keep the common denominator.

Let $\frac{p}{q}$ and $\frac{r}{q}$ be two rational numbers with the same denominator $q$, where $p, q, r$ are integers and $q \neq 0$.

Their sum is given by:

$\frac{p}{q} + \frac{r}{q} = \frac{p+r}{q}$

... (i)

In this formula, we are adding the numerators $p$ and $r$ according to the rules of integer addition, and the denominator remains $q$.

Example 1. Add $\frac{2}{7}$ and $\frac{3}{7}$.

Answer:

Given rational numbers are $\frac{2}{7}$ and $\frac{3}{7}$.

Both numbers have the same denominator, 7.

Using the formula $\frac{p}{q} + \frac{r}{q} = \frac{p+r}{q}$:

$\frac{2}{7} + \frac{3}{7} = \frac{2+3}{7}$

$= \frac{5}{7}$

The sum is $\frac{5}{7}$.

Example 2. Find the sum of $\frac{-5}{11}$ and $\frac{4}{11}$.

Answer:

Given rational numbers are $\frac{-5}{11}$ and $\frac{4}{11}$.

Both numbers have the same denominator, 11.

Using the formula $\frac{p}{q} + \frac{r}{q} = \frac{p+r}{q}$:

$\frac{-5}{11} + \frac{4}{11} = \frac{-5+4}{11}$

Remember the rules for adding integers: $-5 + 4 = -1$.

$= \frac{-1}{11}$

The sum is $\frac{-1}{11}$. This can also be written as $-\frac{1}{11}$.

Case 2: Adding Rational Numbers with Different Denominators

When two or more rational numbers have different denominators (called unlike rational numbers), we cannot directly add the numerators. First, we need to convert them into equivalent rational numbers with the same denominator. The easiest way to find a common denominator is by using the Least Common Multiple (LCM) of the original denominators.

Steps to add rational numbers with different denominators:

Find the LCM of the denominators of the given rational numbers. This LCM will be the new common denominator.
Rewrite each rational number as an equivalent rational number with the LCM as the denominator. To do this, multiply both the numerator and the denominator of each rational number by the factor that makes the denominator equal to the LCM.
Once all rational numbers have the same denominator (the LCM), add them as in Case 1 (add the numerators and keep the common denominator).
Simplify the resulting rational number to its standard form, if necessary.

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where $b \neq d$. Let $L = \text{LCM}(b, d)$. Rewrite $\frac{a}{b} = \frac{a \times \frac{L}{b}}{b \times \frac{L}{b}} = \frac{a'}{L}$ and $\frac{c}{d} = \frac{c \times \frac{L}{d}}{d \times \frac{L}{d}} = \frac{c'}{L}$. Then, $\frac{a}{b} + \frac{c}{d} = \frac{a'}{L} + \frac{c'}{L} = \frac{a'+c'}{L}$.

Example 3. Add $\frac{2}{3}$ and $\frac{1}{4}$.

Answer:

Given rational numbers are $\frac{2}{3}$ and $\frac{1}{4}$.

The denominators are 3 and 4.

Step 1: Find the LCM of 3 and 4.

Since 3 and 4 are coprime (HCF is 1), their LCM is their product: $3 \times 4 = 12$. The new common denominator is 12.

Step 2: Rewrite each rational number with denominator 12.

For $\frac{2}{3}$, we need to multiply the denominator 3 by 4 to get 12. So, multiply the numerator also by 4:

$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

For $\frac{1}{4}$, we need to multiply the denominator 4 by 3 to get 12. So, multiply the numerator also by 3:

$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

Step 3: Add the equivalent rational numbers with the same denominator.

$\frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12}$

$= \frac{8+3}{12}$

$= \frac{11}{12}$

Step 4: Simplify (if needed). 11 and 12 have no common factors other than 1, and the denominator is positive. So, $\frac{11}{12}$ is in standard form.

The sum is $\frac{11}{12}$.

Example 4. Add $\frac{-3}{8}$ and $\frac{5}{12}$.

Answer:

Given rational numbers are $\frac{-3}{8}$ and $\frac{5}{12}$.

The denominators are 8 and 12.

Step 1: Find the LCM of 8 and 12.

Using prime factorisation:

$\begin{array}{c|cc} 2 & 8 \;, & 12 \\ \hline 2 & 4 \; , & 6 \\ \hline 2 & 2 \; , & 3 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$

LCM of 8 and 12 is $2 \times 2 \times 2 \times 3 = 24$. The new common denominator is 24.

Step 2: Rewrite each rational number with denominator 24.

For $\frac{-3}{8}$, we need to multiply the denominator 8 by 3 to get 24. So, multiply the numerator also by 3:

$\frac{-3}{8} = \frac{-3 \times 3}{8 \times 3} = \frac{-9}{24}$

For $\frac{5}{12}$, we need to multiply the denominator 12 by 2 to get 24. So, multiply the numerator also by 2:

$\frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24}$

Step 3: Add the equivalent rational numbers with the same denominator.

$\frac{-3}{8} + \frac{5}{12} = \frac{-9}{24} + \frac{10}{24}$

$= \frac{-9+10}{24}$

Remember the rules for adding integers: $-9 + 10 = 1$.

$= \frac{1}{24}$

Step 4: Simplify (if needed). 1 and 24 have no common factors other than 1, and the denominator is positive. So, $\frac{1}{24}$ is in standard form.

The sum is $\frac{1}{24}$.

Properties of Addition of Rational Numbers

Rational numbers behave nicely under addition, just like integers and whole numbers. They follow certain properties:

1. Closure Property:

When you add any two rational numbers, the result is always another rational number.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, their sum $\frac{a}{b} + \frac{c}{d}$ is also a rational number.

Example: $\frac{1}{2}$ (rational) + $\frac{1}{3}$ (rational). LCM of 2 and 3 is 6. $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{3+2}{6} = \frac{5}{6}$. Since 5 and 6 are integers and $6 \neq 0$, $\frac{5}{6}$ is a rational number. This confirms the closure property.

2. Commutative Property:

The order in which you add two rational numbers does not change the sum. You can swap the positions of the numbers being added.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\frac{a}{b} + \frac{c}{d} = \frac{c}{d} + \frac{a}{b}$.

Example: Let's check with $\frac{-2}{3}$ and $\frac{5}{7}$. LCM of 3 and 7 is 21.

$\frac{-2}{3} + \frac{5}{7} = \frac{-2 \times 7}{3 \times 7} + \frac{5 \times 3}{7 \times 3} = \frac{-14}{21} + \frac{15}{21} = \frac{-14+15}{21} = \frac{1}{21}$

Now, let's swap the order:

$\frac{5}{7} + \frac{-2}{3} = \frac{5 \times 3}{7 \times 3} + \frac{-2 \times 7}{3 \times 7} = \frac{15}{21} + \frac{-14}{21} = \frac{15+(-14)}{21} = \frac{15-14}{21} = \frac{1}{21}$

Since both results are $\frac{1}{21}$, we see that $\frac{-2}{3} + \frac{5}{7} = \frac{5}{7} + \frac{-2}{3}$. This property holds true for all rational numbers.

3. Associative Property:

When adding three or more rational numbers, the way you group them using parentheses does not affect the final sum. You can group the first two numbers or the last two numbers.

For any three rational numbers $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$, $(\frac{a}{b} + \frac{c}{d}) + \frac{e}{f} = \frac{a}{b} + (\frac{c}{d} + \frac{e}{f})$.

Example: Let's add $\frac{1}{2}$, $\frac{-1}{3}$, and $\frac{1}{4}$. LCM of 2, 3, and 4 is 12.

LHS: $(\frac{1}{2} + \frac{-1}{3}) + \frac{1}{4}$

$\frac{1}{2} + \frac{-1}{3} = \frac{1 \times 6}{2 \times 6} + \frac{-1 \times 4}{3 \times 4} = \frac{6}{12} + \frac{-4}{12} = \frac{6+(-4)}{12} = \frac{2}{12} = \frac{1}{6}$

So, $(\frac{1}{2} + \frac{-1}{3}) + \frac{1}{4} = \frac{1}{6} + \frac{1}{4}$

LCM of 6 and 4 is 12.

$\frac{1}{6} + \frac{1}{4} = \frac{1 \times 2}{6 \times 2} + \frac{1 \times 3}{4 \times 3} = \frac{2}{12} + \frac{3}{12} = \frac{2+3}{12} = \frac{5}{12}$

RHS: $\frac{1}{2} + (\frac{-1}{3} + \frac{1}{4})$

$\frac{-1}{3} + \frac{1}{4} = \frac{-1 \times 4}{3 \times 4} + \frac{1 \times 3}{4 \times 3} = \frac{-4}{12} + \frac{3}{12} = \frac{-4+3}{12} = \frac{-1}{12}$

So, $\frac{1}{2} + (\frac{-1}{3} + \frac{1}{4}) = \frac{1}{2} + \frac{-1}{12}$

LCM of 2 and 12 is 12.

$\frac{1}{2} + \frac{-1}{12} = \frac{1 \times 6}{2 \times 6} + \frac{-1}{12} = \frac{6}{12} + \frac{-1}{12} = \frac{6+(-1)}{12} = \frac{6-1}{12} = \frac{5}{12}$

Since LHS = RHS = $\frac{5}{12}$, the associative property is verified.

4. Additive Identity:

There exists a special rational number, 0, which when added to any rational number, leaves the rational number unchanged. This number 0 is called the additive identity for rational numbers.

For any rational number $\frac{a}{b}$, $\frac{a}{b} + 0 = 0 + \frac{a}{b} = \frac{a}{b}$.

We can write $0$ as $\frac{0}{1}$, $\frac{0}{2}$, etc. Example: $\frac{3}{5} + 0 = \frac{3}{5} + \frac{0}{5} = \frac{3+0}{5} = \frac{3}{5}$. $0 + \frac{-7}{9} = \frac{0}{9} + \frac{-7}{9} = \frac{0+(-7)}{9} = \frac{-7}{9}$.

5. Additive Inverse (Negative of a Rational Number):

For every rational number $\frac{a}{b}$, there exists another rational number, denoted by $-\frac{a}{b}$, such that when they are added together, the sum is the additive identity (0). This number $-\frac{a}{b}$ is called the additive inverse or the negative of $\frac{a}{b}$. Similarly, $\frac{a}{b}$ is the additive inverse of $-\frac{a}{b}$.

For any rational number $\frac{a}{b}$, $\frac{a}{b} + (-\frac{a}{b}) = 0$.

The additive inverse of $\frac{a}{b}$ is $\frac{-a}{b}$ or $\frac{a}{-b}$ (since $\frac{-a}{b} = \frac{a}{-b}$).

Example: The additive inverse of $\frac{2}{3}$ is $-\frac{2}{3}$. Check: $\frac{2}{3} + (-\frac{2}{3}) = \frac{2-2}{3} = \frac{0}{3} = 0$.

The additive inverse of $-\frac{5}{7}$ is $\frac{5}{7}$. Check: $-\frac{5}{7} + \frac{5}{7} = \frac{-5+5}{7} = \frac{0}{7} = 0$.

The additive inverse of 0 is 0 itself, since $0 + 0 = 0$.

To find the additive inverse of a rational number, simply change the sign of the numerator (or the entire fraction).

Subtraction of Rational Numbers

Subtraction of rational numbers is closely related to addition. Subtracting a rational number is the same as adding its additive inverse (or negative).

If $\frac{a}{b}$ and $\frac{c}{d}$ are two rational numbers, then subtracting $\frac{c}{d}$ from $\frac{a}{b}$ is defined as adding the additive inverse of $\frac{c}{d}$ to $\frac{a}{b}$.

So, the general rule for subtraction is:

$\frac{a}{b} - \frac{c}{d} = \frac{a}{b} + (\text{additive inverse of } \frac{c}{d})$

Since the additive inverse of $\frac{c}{d}$ is $-\frac{c}{d}$ (or $\frac{-c}{d}$ or $\frac{c}{-d}$), we have:

$\frac{a}{b} - \frac{c}{d} = \frac{a}{b} + \frac{-c}{d}$

... (i)

Using this definition, the procedure for subtraction mirrors that of addition, considering two cases based on the denominators.

Case 1: Subtracting Rational Numbers with the Same Denominators

To subtract rational numbers with the same denominator, we subtract the numerator of the second number from the numerator of the first number and keep the common denominator.

If $\frac{p}{q}$ and $\frac{r}{q}$ are two rational numbers with the same non-zero denominator $q$, then:

$\frac{p}{q} - \frac{r}{q} = \frac{p-r}{q}$

... (ii)

This comes directly from the definition of subtraction using additive inverse:

$\frac{p}{q} - \frac{r}{q} = \frac{p}{q} + (\frac{-r}{q}) = \frac{p+(-r)}{q} = \frac{p-r}{q}$

Example 1. Subtract $\frac{3}{5}$ from $\frac{7}{5}$.

Answer:

We need to calculate $\frac{7}{5} - \frac{3}{5}$.

Both rational numbers have the same denominator, 5.

Using the formula $\frac{p}{q} - \frac{r}{q} = \frac{p-r}{q}$:

$\frac{7}{5} - \frac{3}{5} = \frac{7-3}{5}$

$= \frac{4}{5}$

The difference is $\frac{4}{5}$.

Example 2. Subtract $\frac{-4}{9}$ from $\frac{2}{9}$.

Answer:

We need to calculate $\frac{2}{9} - (\frac{-4}{9})$.

The denominators are the same (9). We can use the formula $\frac{p}{q} - \frac{r}{q} = \frac{p-r}{q}$, where $p=2$, $q=9$, and $r=-4$.

$\frac{2}{9} - (\frac{-4}{9}) = \frac{2 - (-4)}{9}$

Remember that subtracting a negative number is the same as adding the positive number: $2 - (-4) = 2 + 4 = 6$.

$= \frac{6}{9}$

The result $\frac{6}{9}$ is a rational number. Let's reduce it to its standard form by dividing the numerator and denominator by their HCF, which is 3.

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

Alternatively, using the additive inverse concept:

$\frac{2}{9} - (\frac{-4}{9}) = \frac{2}{9} + (\text{additive inverse of } \frac{-4}{9})$

The additive inverse of $\frac{-4}{9}$ is $-(\frac{-4}{9}) = \frac{4}{9}$.

$= \frac{2}{9} + \frac{4}{9} = \frac{2+4}{9} = \frac{6}{9} = \frac{2}{3}$

The difference is $\frac{2}{3}$.

Case 2: Subtracting Rational Numbers with Different Denominators

When rational numbers have different denominators, we first convert them to equivalent rational numbers with a common denominator, usually the LCM of the original denominators. Then, we subtract them as in Case 1.

Steps to subtract rational numbers with different denominators:

Find the LCM of the denominators.
Rewrite each rational number as an equivalent rational number with the LCM as the denominator.
Subtract the numerators of the equivalent rational numbers and keep the common denominator.
Simplify the result to its standard form, if necessary.

Example 3. Subtract $\frac{3}{4}$ from $\frac{5}{6}$.

Answer:

We need to calculate $\frac{5}{6} - \frac{3}{4}$.

The denominators are 6 and 4.

Step 1: Find the LCM of 6 and 4.

Multiples of 6: 6, 12, 18, ...

Multiples of 4: 4, 8, 12, 16, ...

LCM(6, 4) = 12. The common denominator is 12.

Step 2: Rewrite the rational numbers with denominator 12.

For $\frac{5}{6}$, multiply numerator and denominator by $12 \div 6 = 2$:

$\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$

For $\frac{3}{4}$, multiply numerator and denominator by $12 \div 4 = 3$:

$\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

Step 3: Subtract the equivalent rational numbers.

$\frac{5}{6} - \frac{3}{4} = \frac{10}{12} - \frac{9}{12}$

$= \frac{10-9}{12}$

$= \frac{1}{12}$

Step 4: Simplify (if needed). $\frac{1}{12}$ is already in standard form.

The difference is $\frac{1}{12}$.

Example 4. Subtract $\frac{-2}{7}$ from $\frac{-5}{8}$.

Answer:

We need to calculate $\frac{-5}{8} - (\frac{-2}{7})$.

Using the definition of subtraction (adding the additive inverse):

$\frac{-5}{8} - (\frac{-2}{7}) = \frac{-5}{8} + (\text{additive inverse of } \frac{-2}{7})$

The additive inverse of $\frac{-2}{7}$ is $-(\frac{-2}{7}) = \frac{2}{7}$.

$= \frac{-5}{8} + \frac{2}{7}$

Now we need to add $\frac{-5}{8}$ and $\frac{2}{7}$. The denominators are 8 and 7.

Step 1: Find the LCM of 8 and 7. Since 8 and 7 are coprime, their LCM is $8 \times 7 = 56$. The common denominator is 56.

Step 2: Rewrite the rational numbers with denominator 56.

For $\frac{-5}{8}$, multiply numerator and denominator by $56 \div 8 = 7$:

$\frac{-5}{8} = \frac{-5 \times 7}{8 \times 7} = \frac{-35}{56}$

For $\frac{2}{7}$, multiply numerator and denominator by $56 \div 7 = 8$:

$\frac{2}{7} = \frac{2 \times 8}{7 \times 8} = \frac{16}{56}$

Step 3: Add the equivalent rational numbers.

$\frac{-35}{56} + \frac{16}{56} = \frac{-35+16}{56}$

Add the integers in the numerator: $-35 + 16 = -19$.

$= \frac{-19}{56}$

Step 4: Simplify (if needed). -19 and 56 have no common factors other than 1, and the denominator is positive. So, $\frac{-19}{56}$ is in standard form.

The difference is $\frac{-19}{56}$.

Properties of Subtraction of Rational Numbers

Let's examine the properties of subtraction for rational numbers.

1. Closure Property:

The difference between any two rational numbers is always a rational number.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\frac{a}{b} - \frac{c}{d}$ is also a rational number.

Example: Consider $\frac{1}{2}$ and $\frac{1}{3}$. Their difference is $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$. Since 1 and 6 are integers and $6 \neq 0$, $\frac{1}{6}$ is a rational number. This property holds true for all pairs of rational numbers.

2. Commutative Property:

Subtraction of rational numbers is not commutative. This means changing the order of the numbers in a subtraction problem changes the result.

For any two distinct rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\frac{a}{b} - \frac{c}{d} \neq \frac{c}{d} - \frac{a}{b}$.

Example: Consider $\frac{1}{2}$ and $\frac{1}{3}$.

$\frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$

$\frac{1}{3} - \frac{1}{2} = \frac{2-3}{6} = \frac{-1}{6}$

Since $\frac{1}{6} \neq \frac{-1}{6}$, $\frac{1}{2} - \frac{1}{3} \neq \frac{1}{3} - \frac{1}{2}$. Subtraction is not commutative for rational numbers.

3. Associative Property:

Subtraction of rational numbers is not associative. This means when subtracting three or more rational numbers, the way you group them affects the result.

For any three distinct rational numbers $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$, $(\frac{a}{b} - \frac{c}{d}) - \frac{e}{f} \neq \frac{a}{b} - (\frac{c}{d} - \frac{e}{f})$.

Example: Let's check with $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{4}$.

LHS: $(\frac{1}{2} - \frac{1}{3}) - \frac{1}{4}$

$\frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$

So, $(\frac{1}{2} - \frac{1}{3}) - \frac{1}{4} = \frac{1}{6} - \frac{1}{4}$

LCM of 6 and 4 is 12.

$\frac{1}{6} - \frac{1}{4} = \frac{1 \times 2}{6 \times 2} - \frac{1 \times 3}{4 \times 3} = \frac{2}{12} - \frac{3}{12} = \frac{2-3}{12} = \frac{-1}{12}$

RHS: $\frac{1}{2} - (\frac{1}{3} - \frac{1}{4})$

$\frac{1}{3} - \frac{1}{4} = \frac{4-3}{12} = \frac{1}{12}$

So, $\frac{1}{2} - (\frac{1}{3} - \frac{1}{4}) = \frac{1}{2} - \frac{1}{12}$

LCM of 2 and 12 is 12.

$\frac{1}{2} - \frac{1}{12} = \frac{1 \times 6}{2 \times 6} - \frac{1}{12} = \frac{6}{12} - \frac{1}{12} = \frac{6-1}{12} = \frac{5}{12}$

Since LHS = $\frac{-1}{12}$ and RHS = $\frac{5}{12}$, LHS $\neq$ RHS. Subtraction is not associative for rational numbers.

Multiplication of Rational Numbers

Multiplication of rational numbers is a straightforward operation. To multiply two rational numbers, you simply multiply their numerators together and multiply their denominators together.

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where $a, b, c, d$ are integers and $b \neq 0, d \neq 0$.

Their product is given by the formula:

$\frac{a}{b} \times \frac{c}{d} = \frac{\text{Product of numerators}}{\text{Product of denominators}} = \frac{a \times c}{b \times d}$

... (i)

The resulting fraction $\frac{a \times c}{b \times d}$ is a rational number because $a \times c$ is an integer (product of integers is an integer), and $b \times d$ is a non-zero integer (product of non-zero integers is a non-zero integer).

Example 1. Multiply $\frac{2}{3}$ by $\frac{4}{5}$.

Answer:

Given rational numbers: $\frac{2}{3}$ and $\frac{4}{5}$.

Using the multiplication rule:

$\frac{2}{3} \times \frac{4}{5} = \frac{2 \times 4}{3 \times 5}$

$= \frac{8}{15}$

The product is $\frac{8}{15}$. This is in standard form as HCF(8, 15) = 1 and the denominator is positive.

Example 2. Find the product of $\frac{-3}{7}$ and $\frac{5}{8}$.

Answer:

Given rational numbers: $\frac{-3}{7}$ and $\frac{5}{8}$.

Using the multiplication rule:

$\frac{-3}{7} \times \frac{5}{8} = \frac{-3 \times 5}{7 \times 8}$

Multiplying integers in the numerator and denominator:

$= \frac{-15}{56}$

The product is $\frac{-15}{56}$. This is in standard form as HCF(15, 56) = 1 and the denominator is positive.

Example 3. Multiply $\frac{-4}{5}$ by $\frac{-10}{12}$.

Answer:

Given rational numbers: $\frac{-4}{5}$ and $\frac{-10}{12}$.

Using the multiplication rule:

$\frac{-4}{5} \times \frac{-10}{12} = \frac{(-4) \times (-10)}{5 \times 12}$

Multiplying integers: $(-4) \times (-10) = 40$ and $5 \times 12 = 60$.

$= \frac{40}{60}$

The product is $\frac{40}{60}$. This can be simplified to its standard form. The HCF of 40 and 60 is 20.

$\frac{40}{60} = \frac{40 \div 20}{60 \div 20} = \frac{2}{3}$

The standard form of the product is $\frac{2}{3}$.

Alternative Method (Cancelling Common Factors before Multiplication):

You can often simplify the multiplication by cancelling common factors between any numerator and any denominator before multiplying.

$\frac{-4}{5} \times \frac{-10}{12} = \frac{-\cancel{4}^{\normalsize 1}}{ \cancel{5}_{\normalsize 1}} \times \frac{-\cancel{10}^{\normalsize 2}}{ \cancel{12}_{\normalsize 3}}$

Here, we cancelled 4 with 12 (4 divides both, $4 \div 4 = 1$, $12 \div 4 = 3$) and 5 with 10 (5 divides both, $5 \div 5 = 1$, $10 \div 5 = 2$).

Now multiply the simplified numerators and denominators:

$= \frac{-1}{1} \times \frac{-2}{3}$

$= \frac{(-1) \times (-2)}{1 \times 3} = \frac{2}{3}$

This method often makes the calculation easier and directly gives the result in standard form.

Properties of Multiplication of Rational Numbers

Multiplication of rational numbers also satisfies several important properties:

1. Closure Property:

The product of any two rational numbers is always a rational number.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, their product $\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$ is also a rational number (since $ac$ and $bd$ are integers, and $bd \neq 0$ if $b \neq 0$ and $d \neq 0$).

Example: $\frac{1}{2}$ (rational) $\times \frac{1}{3}$ (rational) $= \frac{1 \times 1}{2 \times 3} = \frac{1}{6}$. $\frac{1}{6}$ is a rational number. This confirms closure under multiplication.

2. Commutative Property:

The order in which you multiply two rational numbers does not change the product.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b}$.

Example: Let's check with $\frac{-2}{3}$ and $\frac{5}{7}$.

$\frac{-2}{3} \times \frac{5}{7} = \frac{-2 \times 5}{3 \times 7} = \frac{-10}{21}$

$\frac{5}{7} \times \frac{-2}{3} = \frac{5 \times (-2)}{7 \times 3} = \frac{-10}{21}$

Since both results are $\frac{-10}{21}$, we see that $\frac{-2}{3} \times \frac{5}{7} = \frac{5}{7} \times \frac{-2}{3}$. Multiplication of rational numbers is commutative.

3. Associative Property:

When multiplying three or more rational numbers, the grouping of the numbers using parentheses does not affect the final product.

For any three rational numbers $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$, $(\frac{a}{b} \times \frac{c}{d}) \times \frac{e}{f} = \frac{a}{b} \times (\frac{c}{d} \times \frac{e}{f})$.

Example: Let's multiply $\frac{1}{2}$, $\frac{-1}{3}$, and $\frac{1}{4}$.

LHS: $(\frac{1}{2} \times \frac{-1}{3}) \times \frac{1}{4}$

$\frac{1}{2} \times \frac{-1}{3} = \frac{1 \times (-1)}{2 \times 3} = \frac{-1}{6}$

So, $(\frac{1}{2} \times \frac{-1}{3}) \times \frac{1}{4} = \frac{-1}{6} \times \frac{1}{4} = \frac{-1 \times 1}{6 \times 4} = \frac{-1}{24}$

RHS: $\frac{1}{2} \times (\frac{-1}{3} \times \frac{1}{4})$

$\frac{-1}{3} \times \frac{1}{4} = \frac{-1 \times 1}{3 \times 4} = \frac{-1}{12}$

So, $\frac{1}{2} \times (\frac{-1}{3} \times \frac{1}{4}) = \frac{1}{2} \times \frac{-1}{12} = \frac{1 \times (-1)}{2 \times 12} = \frac{-1}{24}$

Since LHS = RHS = $\frac{-1}{24}$, the associative property is verified for multiplication of rational numbers.

4. Multiplicative Identity (Role of 1):

There exists a special rational number, 1, which when multiplied by any rational number, leaves the rational number unchanged. This number 1 is called the multiplicative identity for rational numbers.

For any rational number $\frac{a}{b}$, $\frac{a}{b} \times 1 = 1 \times \frac{a}{b} = \frac{a}{b}$.

We can write $1$ as $\frac{1}{1}$, $\frac{2}{2}$, $\frac{-5}{-5}$, etc. Example: $\frac{3}{5} \times 1 = \frac{3}{5} \times \frac{1}{1} = \frac{3 \times 1}{5 \times 1} = \frac{3}{5}$. $1 \times \frac{-7}{9} = \frac{1}{1} \times \frac{-7}{9} = \frac{1 \times (-7)}{1 \times 9} = \frac{-7}{9}$.

5. Multiplicative Inverse (Reciprocal):

For every non-zero rational number $\frac{a}{b}$, there exists another rational number, denoted by its reciprocal or multiplicative inverse, such that when multiplied together, the product is the multiplicative identity (1).

If $\frac{a}{b}$ is a non-zero rational number (meaning $a \neq 0$ and $b \neq 0$), its multiplicative inverse is $\frac{b}{a}$.

This is because:

$\frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = \frac{ab}{ab} = 1$

To find the reciprocal of a non-zero rational number, simply swap the numerator and the denominator.

Example: The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$. Check: $\frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = 1$.

The reciprocal of $-\frac{5}{7}$ (which is $\frac{-5}{7}$) is $\frac{7}{-5}$, which is equivalent to $-\frac{7}{5}$. Check: $\frac{-5}{7} \times \frac{7}{-5} = \frac{-35}{-35} = 1$.

The reciprocal of an integer, say 4 (which is $\frac{4}{1}$), is $\frac{1}{4}$.

Important Note: The number 0 does not have a reciprocal. This is because if 0 had a reciprocal, say $x$, then $0 \times x$ should equal 1. But we know that $0 \times x = 0$ for any value of $x$. Since $0 \neq 1$, there is no such number $x$ that can be the reciprocal of 0.

6. Distributive Property of Multiplication over Addition and Subtraction:

Multiplication of rational numbers is distributive over both addition and subtraction. This property connects multiplication and addition/subtraction.

For any three rational numbers $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$:

$\frac{a}{b} \times (\frac{c}{d} + \frac{e}{f}) = (\frac{a}{b} \times \frac{c}{d}) + (\frac{a}{b} \times \frac{e}{f})$ (Distributive property over addition)

... (ii)

$\frac{a}{b} \times (\frac{c}{d} - \frac{e}{f}) = (\frac{a}{b} \times \frac{c}{d}) - (\frac{a}{b} \times \frac{e}{f})$ (Distributive property over subtraction)

... (iii)

Example: Let's verify property (ii) with $\frac{1}{2} \times (\frac{1}{3} + \frac{1}{4})$.

LHS: $\frac{1}{2} \times (\frac{1}{3} + \frac{1}{4})$. First calculate the sum inside the parenthesis. LCM of 3 and 4 is 12.

$\frac{1}{3} + \frac{1}{4} = \frac{1 \times 4}{3 \times 4} + \frac{1 \times 3}{4 \times 3} = \frac{4}{12} + \frac{3}{12} = \frac{4+3}{12} = \frac{7}{12}$

Now, $\frac{1}{2} \times (\frac{1}{3} + \frac{1}{4}) = \frac{1}{2} \times \frac{7}{12} = \frac{1 \times 7}{2 \times 12} = \frac{7}{24}$

RHS: $(\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{2} \times \frac{1}{4})$. First calculate the products in the parentheses.

$\frac{1}{2} \times \frac{1}{3} = \frac{1 \times 1}{2 \times 3} = \frac{1}{6}$

$\frac{1}{2} \times \frac{1}{4} = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}$

Now, add the products: $\frac{1}{6} + \frac{1}{8}$. LCM of 6 and 8 is 24.

$\frac{1}{6} + \frac{1}{8} = \frac{1 \times 4}{6 \times 4} + \frac{1 \times 3}{8 \times 3} = \frac{4}{24} + \frac{3}{24} = \frac{4+3}{24} = \frac{7}{24}$

Since LHS = $\frac{7}{24}$ and RHS = $\frac{7}{24}$, LHS = RHS. The distributive property of multiplication over addition is verified.

7. Role of Zero in Multiplication:

When any rational number is multiplied by zero, the product is always zero.

For any rational number $\frac{a}{b}$, $\frac{a}{b} \times 0 = 0 \times \frac{a}{b} = 0$.

Example: $\frac{3}{5} \times 0 = \frac{3}{5} \times \frac{0}{1} = \frac{3 \times 0}{5 \times 1} = \frac{0}{5} = 0$.

Division of Rational Numbers

Division of rational numbers is defined in terms of multiplication. Dividing by a rational number (that is not zero) is the same as multiplying by its multiplicative inverse (reciprocal).

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where $a, b, c, d$ are integers and $b \neq 0, d \neq 0$. For division, we also have the crucial condition that the divisor, $\frac{c}{d}$, must not be zero. This means $c$ cannot be zero ($c \neq 0$).

To divide $\frac{a}{b}$ by $\frac{c}{d}$ (where $\frac{c}{d} \neq 0$), we use the rule:

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times (\text{multiplicative inverse of } \frac{c}{d})$

... (i)

We know that the multiplicative inverse (reciprocal) of $\frac{c}{d}$ is $\frac{d}{c}$ (since $\frac{c}{d} \neq 0$, $c \neq 0$ and $d \neq 0$ are implicit if $\frac{c}{d}$ is in standard form or can be written as such; if $c=0$, the original number is 0, which cannot be a divisor). So, we can write:

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

... (ii)

Now, the division problem is converted into a multiplication problem, which we already know how to solve. We multiply the numerators and multiply the denominators.

Example 1. Divide $\frac{2}{3}$ by $\frac{4}{5}$.

Answer:

Given rational numbers: $\frac{2}{3}$ and $\frac{4}{5}$. We need to calculate $\frac{2}{3} \div \frac{4}{5}$.

Step 1: Identify the first rational number ($\frac{a}{b}$) and the second rational number (divisor) ($\frac{c}{d}$).

First number = $\frac{2}{3}$

Second number = $\frac{4}{5}$

The second number $\frac{4}{5}$ is not zero, so division is possible.

Step 2: Find the reciprocal of the second rational number.

The reciprocal of $\frac{4}{5}$ is $\frac{5}{4}$.

Step 3: Multiply the first rational number by the reciprocal of the second rational number.

$\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4}$

Now, perform the multiplication:

$= \frac{2 \times 5}{3 \times 4} = \frac{10}{12}$

Step 4: Simplify the result to its standard form.

The rational number is $\frac{10}{12}$. The HCF of 10 and 12 is 2.

$\frac{10}{12} = \frac{10 \div 2}{12 \div 2} = \frac{5}{6}$

The quotient is $\frac{5}{6}$.

Alternative Method (Cancelling before multiplying):

$\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4}$

Cancel the common factor 2 between numerator 2 and denominator 4:

$= \frac{\cancel{2}^{\normalsize 1}}{3} \times \frac{5}{\cancel{4}_{\normalsize 2}} = \frac{1}{3} \times \frac{5}{2}$

Multiply the simplified terms:

$= \frac{1 \times 5}{3 \times 2} = \frac{5}{6}$

This gives the result directly in standard form.

Example 2. Divide $\frac{-6}{11}$ by $\frac{-18}{22}$.

Answer:

Given rational numbers: $\frac{-6}{11}$ and $\frac{-18}{22}$. We need to calculate $\frac{-6}{11} \div \frac{-18}{22}$.

Step 1: Identify the first rational number and the divisor. Divisor is $\frac{-18}{22}$. Is it zero? $\frac{-18}{22}$ is equivalent to $\frac{-9}{11}$, which is not zero. So division is possible.

Step 2: Find the reciprocal of the divisor $\frac{-18}{22}$.

Reciprocal of $\frac{-18}{22}$ is $\frac{22}{-18}$. Note that $\frac{22}{-18}$ is equivalent to $\frac{-22}{18}$ or $-\frac{11}{9}$. We can use any form, but keeping it as $\frac{22}{-18}$ might be easier for cancellation.

Step 3: Multiply the first rational number by the reciprocal of the second rational number.

$\frac{-6}{11} \div \frac{-18}{22} = \frac{-6}{11} \times \frac{22}{-18}$

Step 4: Perform the multiplication and simplify.

Cancel common factors: 6 with 18 (HCF is 6), 11 with 22 (HCF is 11).

$= \frac{-\cancel{6}^{\normalsize 1}}{\cancel{11}_{\normalsize 1}} \times \frac{\cancel{22}^{\normalsize 2}}{-\cancel{18}_{\normalsize 3}} = \frac{-1}{1} \times \frac{2}{-3}$

Multiply the simplified terms:

$= \frac{(-1) \times 2}{1 \times (-3)} = \frac{-2}{-3}$

A fraction with negative signs in both numerator and denominator is positive:

$= \frac{2}{3}$

The quotient is $\frac{2}{3}$. This is in standard form.

Properties of Division of Rational Numbers

Let's look at the properties of division for rational numbers.

1. Closure Property:

Division of rational numbers is closed for all rational numbers, except for division by zero. This means if you divide a rational number by any non-zero rational number, the result is always a rational number.

For any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, where $\frac{c}{d} \neq 0$, the quotient $\frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}$ is a rational number (since $ad$ and $bc$ are integers, and $bc \neq 0$ because $b \neq 0, c \neq 0, d \neq 0$).

Division by Zero is Undefined: Dividing any number by zero is not possible in mathematics. If we tried to define $\frac{a}{b} \div 0$, using the rule it would be $\frac{a}{b} \times \text{reciprocal of } 0$. But 0 has no reciprocal. Therefore, division by zero is undefined for rational numbers (and indeed for all real numbers).

2. Commutative Property:

Division of rational numbers is not commutative. Changing the order of the numbers being divided changes the result.

For any two distinct non-zero rational numbers $\frac{a}{b}$ and $\frac{c}{d}$, $\frac{a}{b} \div \frac{c}{d} \neq \frac{c}{d} \div \frac{a}{b}$.

Example: Consider $\frac{1}{2}$ and $\frac{1}{3}$.

$\frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}$

$\frac{1}{3} \div \frac{1}{2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}$

Since $\frac{3}{2} \neq \frac{2}{3}$, division is not commutative for rational numbers.

3. Associative Property:

Division of rational numbers is not associative. The way you group three or more rational numbers in a division problem affects the result.

For any three distinct non-zero rational numbers $\frac{a}{b}$, $\frac{c}{d}$, and $\frac{e}{f}$, $(\frac{a}{b} \div \frac{c}{d}) \div \frac{e}{f} \neq \frac{a}{b} \div (\frac{c}{d} \div \frac{e}{f})$.

Example: Let's check with $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{4}$.

LHS: $(\frac{1}{2} \div \frac{1}{3}) \div \frac{1}{4}$

$\frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2}$

So, $(\frac{1}{2} \div \frac{1}{3}) \div \frac{1}{4} = \frac{3}{2} \div \frac{1}{4} = \frac{3}{2} \times \frac{4}{1} = \frac{12}{2} = 6$

RHS: $\frac{1}{2} \div (\frac{1}{3} \div \frac{1}{4})$

$\frac{1}{3} \div \frac{1}{4} = \frac{1}{3} \times \frac{4}{1} = \frac{4}{3}$

So, $\frac{1}{2} \div (\frac{1}{3} \div \frac{1}{4}) = \frac{1}{2} \div \frac{4}{3} = \frac{1}{2} \times \frac{3}{4} = \frac{1 \times 3}{2 \times 4} = \frac{3}{8}$

Since LHS = 6 and RHS = $\frac{3}{8}$, LHS $\neq$ RHS. Division is not associative for rational numbers.

4. Division by 1 and -1:

When a rational number is divided by 1, the quotient is the rational number itself.

$\frac{a}{b} \div 1 = \frac{a}{b} \times \frac{1}{1} = \frac{a \times 1}{b \times 1} = \frac{a}{b}$

Example: $\frac{5}{7} \div 1 = \frac{5}{7}$.
When a rational number is divided by -1, the quotient is its additive inverse (the number with the opposite sign).

$\frac{a}{b} \div (-1) = \frac{a}{b} \times \frac{1}{-1} = \frac{a}{b} \times (-1) = \frac{-a}{b}$

Example: $\frac{5}{7} \div (-1) = \frac{-5}{7}$. Also, $\frac{-3}{4} \div (-1) = \frac{-3}{4} \times \frac{1}{-1} = \frac{-3}{4} \times (-1) = \frac{3}{4}$.

Representation of Rational Numbers on the Number Line

You are already familiar with representing integers on a number line. A number line is a visual representation of numbers as points on a straight line. We usually mark 0 at the center, positive integers to the right of 0, and negative integers to the left of 0 at equal distances.

Just like integers, rational numbers can also be uniquely represented as points on the number line. Every point on the number line corresponds to a unique real number, and rational numbers are a part of the real number system. Let's see how to locate rational numbers on a number line.

Representing Positive Rational Numbers ($\frac{p}{q}$, $p > 0, q > 0$)

Positive rational numbers are located on the number line to the right of 0.

Consider a positive rational number $\frac{p}{q}$ where $p$ and $q$ are positive integers.

If $p < q$: The rational number is a proper fraction. It is greater than 0 and less than 1. So, $\frac{p}{q}$ lies between 0 and 1 on the number line.
If $p = q$: The rational number is equal to 1. It is located at the point marked 1 on the number line.
If $p > q$: The rational number is an improper fraction. It is greater than 1. We can convert it into a mixed number, say $k \frac{r}{q}$, where $k$ is the integer part and $\frac{r}{q}$ is the proper fractional part ($0 < \frac{r}{q} < 1$). This means the rational number $\frac{p}{q}$ lies between the integers $k$ and $k+1$.

Steps to Represent $\frac{p}{q}$ (where $p > 0, q > 0$) on the Number Line:

Draw a number line and mark the integer points (0, 1, 2, -1, -2, ...).
Determine which two integers the rational number $\frac{p}{q}$ lies between.
- If $0 < \frac{p}{q} < 1$ (i.e., $p
- If $\frac{p}{q} = k \frac{r}{q}$ (where $k$ is a whole number and $\frac{r}{q}$ is a proper fraction), it's between $k$ and $k+1$.
Divide the segment between these two integers into $q$ equal parts (where $q$ is the denominator).
Starting from the left integer of the segment (0 or $k$), count $p$ parts to the right. The point you reach represents $\frac{p}{q}$. If you are using the mixed number $k \frac{r}{q}$, divide the segment from $k$ to $k+1$ into $q$ parts and count $r$ parts from $k$.

Example 1. Represent $\frac{1}{2}$ on the number line.

Answer:

Given rational number: $\frac{1}{2}$. Here $p=1, q=2$. Both are positive, and $p < q$.

So, $\frac{1}{2}$ lies between 0 and 1.

Draw a number line and mark 0 and 1.

The denominator is 2, so divide the segment between 0 and 1 into 2 equal parts.

Starting from 0, move 1 part to the right (since the numerator is 1). The point reached represents $\frac{1}{2}$.

Example 2. Represent $\frac{3}{4}$ on the number line.

Answer:

Given rational number: $\frac{3}{4}$. Here $p=3, q=4$. Both are positive, and $p < q$.

So, $\frac{3}{4}$ lies between 0 and 1.

Draw a number line and mark 0 and 1.

The denominator is 4, so divide the segment between 0 and 1 into 4 equal parts.

Starting from 0, move 3 parts to the right (since the numerator is 3). The point reached represents $\frac{3}{4}$.

Example 3. Represent $\frac{5}{3}$ on the number line.

Answer:

Given rational number: $\frac{5}{3}$. Here $p=5, q=3$. Both are positive, and $p > q$.

Convert the improper fraction to a mixed number: $\frac{5}{3} = 1 \frac{2}{3}$.

This means the number is 1 plus $\frac{2}{3}$. So, it lies between the integers 1 and $1+1=2$.

Draw a number line and mark 1 and 2.

The denominator of the fractional part is 3, so divide the segment between 1 and 2 into 3 equal parts.

Starting from 1, move 2 parts to the right (since the numerator of the fractional part is 2). The point reached represents $\frac{5}{3}$.

Representing Negative Rational Numbers ($-\frac{p}{q}$, $p > 0, q > 0$)

Negative rational numbers are located on the number line to the left of 0.

To represent a negative rational number like $-\frac{p}{q}$ (where $p > 0, q > 0$), we first consider its absolute value, $\frac{p}{q}$. We find where $\frac{p}{q}$ would be located to the right of 0, and then mark the corresponding point at the same distance to the left of 0.

If $p < q$: The absolute value $\frac{p}{q}$ is between 0 and 1. So, $-\frac{p}{q}$ lies between -1 and 0.
If $p = q$: The absolute value $\frac{p}{q}$ is 1. So, $-\frac{p}{q}$ is -1. It is located at the point marked -1.
If $p > q$: The absolute value $\frac{p}{q} = k \frac{r}{q}$, which is between $k$ and $k+1$. So, $-\frac{p}{q}$ lies between the integers $-(k+1)$ and $-k$. For example, $-\frac{5}{3} = -1 \frac{2}{3}$, which is between -2 and -1.

Steps to Represent $-\frac{p}{q}$ (where $p > 0, q > 0$) on the Number Line:

Draw a number line and mark the integer points.
Determine which two integers the rational number $-\frac{p}{q}$ lies between based on its absolute value $\frac{p}{q}$.
- If $0 < \frac{p}{q} < 1$, it's between -1 and 0.
- If $\frac{p}{q} = k \frac{r}{q}$, it's between $-(k+1)$ and $-k$.
Divide the segment between these two integers into $q$ equal parts (where $q$ is the denominator).
Starting from the right integer of the segment (0 or $-k$), count $p$ parts to the left. The point you reach represents $-\frac{p}{q}$. If you are using the mixed number $-k \frac{r}{q}$, divide the segment from $-k$ to $-(k+1)$ into $q$ parts and count $r$ parts from $-k$ to the left.

Example 4. Represent $-\frac{1}{2}$ on the number line.

Answer:

Given rational number: $-\frac{1}{2}$. The absolute value is $\frac{1}{2}$. $\frac{1}{2}$ is between 0 and 1. So, $-\frac{1}{2}$ is between -1 and 0.

Draw a number line and mark 0 and -1.

The denominator is 2, so divide the segment between -1 and 0 into 2 equal parts.

Starting from 0, move 1 part to the left (since the numerator is 1, and we are representing the negative value). The point reached represents $-\frac{1}{2}$.

Example 5. Represent $-\frac{3}{4}$ on the number line.

Answer:

Given rational number: $-\frac{3}{4}$. The absolute value is $\frac{3}{4}$. $\frac{3}{4}$ is between 0 and 1. So, $-\frac{3}{4}$ is between -1 and 0.

Draw a number line and mark 0 and -1.

The denominator is 4, so divide the segment between -1 and 0 into 4 equal parts.

Starting from 0, move 3 parts to the left (since the numerator is 3). The point reached represents $-\frac{3}{4}$.

Example 6. Represent $-\frac{5}{3}$ on the number line.

Answer:

Given rational number: $-\frac{5}{3}$. The absolute value is $\frac{5}{3}$.

Convert the absolute value to a mixed number: $\frac{5}{3} = 1 \frac{2}{3}$. This is between 1 and 2.

So, $-\frac{5}{3}$ is between -2 and -1.

Draw a number line and mark -1 and -2.

The denominator of the fractional part is 3, so divide the segment between -2 and -1 into 3 equal parts.

Starting from -1 (which is equivalent to $-1 \frac{0}{3}$), move 2 parts to the left (since the numerator of the fractional part is 2). The point reached represents $-\frac{5}{3}$.

Rational Numbers Between Two Rational Numbers

We know that between any two integers, there might be a finite number of integers (e.g., between 1 and 5 are 2, 3, 4) or no integers at all (e.g., between 1 and 2). However, when it comes to rational numbers, the situation is quite different.

An important property of rational numbers is that between any two distinct rational numbers, there exist infinitely many other rational numbers. This property is sometimes referred to as the density property of rational numbers.

Let's explore how to find rational numbers that lie between two given rational numbers.

Method 1: Using the Average (Mean)

One simple way to find a rational number between any two given rational numbers is to calculate their average (arithmetic mean). The average of two numbers always lies between them.

If $a$ and $b$ are two distinct rational numbers, a rational number between $a$ and $b$ can be found using the formula:

$\text{Rational number between } a \text{ and } b = \frac{a+b}{2}$

... (i)

Once you find one rational number between $a$ and $b$, you can use this method again with one of the original numbers and the new number to find another rational number, and so on. You can repeat this process indefinitely to find as many rational numbers as you need between $a$ and $b$, illustrating the infinite number of rational numbers between any two distinct ones.

Example 1. Find a rational number between $\frac{1}{2}$ and $\frac{1}{3}$.

Answer:

Given rational numbers are $a = \frac{1}{2}$ and $b = \frac{1}{3}$.

Using the average method, a rational number between them is:

$\frac{a+b}{2} = \frac{\frac{1}{2} + \frac{1}{3}}{2}$

First, add the rational numbers in the numerator. Find the LCM of the denominators 2 and 3, which is 6.

$\frac{1}{2} + \frac{1}{3} = \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} = \frac{3}{6} + \frac{2}{6} = \frac{3+2}{6} = \frac{5}{6}$

Now, substitute this sum back into the average formula:

$\frac{\frac{5}{6}}{2} = \frac{5}{6} \div 2$

Remember that dividing by a number is the same as multiplying by its reciprocal. The reciprocal of 2 (which is $\frac{2}{1}$) is $\frac{1}{2}$.

$= \frac{5}{6} \times \frac{1}{2} = \frac{5 \times 1}{6 \times 2} = \frac{5}{12}$

So, $\frac{5}{12}$ is a rational number between $\frac{1}{2}$ and $\frac{1}{3}$.

To verify, we can express all three numbers with a common denominator. LCM of 2, 3, and 12 is 12.

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

$\frac{5}{12} = \frac{5}{12}$

$\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$

Comparing the numerators with the same denominator, we have $4 < 5 < 6$. Thus, $\frac{4}{12} < \frac{5}{12} < \frac{6}{12}$, which means $\frac{1}{3} < \frac{5}{12} < \frac{1}{2}$. The answer is correct.

Method 2: Making Denominators Equal (and Finding Equivalent Fractions)

This method is particularly useful when you need to find several rational numbers between two given rational numbers. The idea is to rewrite the given rational numbers as equivalent rational numbers with a sufficiently large common denominator, so that there are integers between their numerators.

Steps for Method 2:

Find a common denominator for the two given rational numbers. The LCM of the denominators is a good choice.
Rewrite both rational numbers as equivalent fractions with this common denominator.
If there are not enough integers between the new numerators to give the required number of rational numbers, multiply both the numerator and the denominator of *both* equivalent fractions by a suitable integer (e.g., 10, or the number of rational numbers required plus one). This creates more "space" between the numerators while maintaining the value of the original rational numbers.
Once you have two equivalent rational numbers with the same denominator and enough integers between their numerators, you can list the rational numbers between them by taking the integers between the numerators and using the common denominator.

Example 2. Find 5 rational numbers between $\frac{1}{2}$ and $\frac{2}{3}$.

Answer:

Given rational numbers: $\frac{1}{2}$ and $\frac{2}{3}$.

Step 1: Find the LCM of the denominators 2 and 3. LCM(2, 3) = 6. The common denominator is 6.

Step 2: Rewrite the numbers with denominator 6.

$\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

Now we need to find 5 rational numbers between $\frac{3}{6}$ and $\frac{4}{6}$. The integers between 3 and 4 are none. We need more space between the numerators.

Step 3: Multiply both the numerator and denominator of $\frac{3}{6}$ and $\frac{4}{6}$ by a suitable integer. We need 5 numbers, so multiplying by 6 or 10 will usually create enough space. Let's multiply by 10:

$\frac{3}{6} = \frac{3 \times 10}{6 \times 10} = \frac{30}{60}$

$\frac{4}{6} = \frac{4 \times 10}{6 \times 10} = \frac{40}{60}$

Now we need to find 5 rational numbers between $\frac{30}{60}$ and $\frac{40}{60}$. The integers between 30 and 40 are 31, 32, 33, 34, 35, 36, 37, 38, 39.

Step 4: Form rational numbers using these integers as numerators and the common denominator 60.

We can choose any 5 of these. Five rational numbers between $\frac{1}{2}$ and $\frac{2}{3}$ are, for example:

$\frac{31}{60}, \frac{32}{60}, \frac{33}{60}, \frac{34}{60}, \frac{35}{60}$

These are just five possible numbers; you could choose any other 5 from the list $\frac{31}{60}, \frac{32}{60}, ..., \frac{39}{60}$.

Example 3. Find 10 rational numbers between $-\frac{2}{5}$ and $\frac{1}{2}$.

Answer:

Given rational numbers: $-\frac{2}{5}$ and $\frac{1}{2}$.

Step 1: Find the LCM of the denominators 5 and 2. LCM(5, 2) = 10. The common denominator is 10.

Step 2: Rewrite the numbers with denominator 10.

$-\frac{2}{5} = \frac{-2 \times 2}{5 \times 2} = \frac{-4}{10}$

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$

Now we need to find 10 rational numbers between $\frac{-4}{10}$ and $\frac{5}{10}$. The integers between -4 and 5 are -3, -2, -1, 0, 1, 2, 3, 4. If we use these as numerators with denominator 10, we get:

$\frac{-3}{10}, \frac{-2}{10}, \frac{-1}{10}, \frac{0}{10}, \frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10}$

There are $4 - (-3) + 1 = 8$ integers between -4 and 5 (exclusive of -4 and 5). So we can only find 8 rational numbers this way using denominator 10.

We need 10 rational numbers. So, we need to create more "space" by multiplying the numerator and denominator by a larger number. We currently have $5 - (-4) - 1 = 8$ integers between the numerators -4 and 5. We need at least 10 integers. Let's try multiplying by 3 (since $3 \times 8 = 24$, more than 10 needed). If we had multiplied by 2, we'd have $5 \times 2 - (-4 \times 2) - 1 = 10 - (-8) - 1 = 10 + 8 - 1 = 17$ integers, which is also enough.

Let's multiply numerator and denominator of $\frac{-4}{10}$ and $\frac{5}{10}$ by 3:

$\frac{-4}{10} = \frac{-4 \times 3}{10 \times 3} = \frac{-12}{30}$

$\frac{5}{10} = \frac{5 \times 3}{10 \times 3} = \frac{15}{30}$

Now we need to find 10 rational numbers between $\frac{-12}{30}$ and $\frac{15}{30}$. The integers between -12 and 15 are -11, -10, -9, ..., 13, 14.

We can choose any 10 of these integers as numerators, keeping the denominator as 30. Some possible rational numbers are:

$\frac{-11}{30}, \frac{-10}{30}, \frac{-9}{30}, \frac{-8}{30}, \frac{-7}{30}, \frac{-6}{30}, \frac{-5}{30}, \frac{-4}{30}, \frac{-3}{30}, \frac{-2}{30}$

You could also choose positive ones, like $\frac{1}{30}, \frac{2}{30}, \dots, \frac{14}{30}$, or a mix of positive and negative numbers. For example:

$\frac{-5}{30}, \frac{-2}{30}, \frac{0}{30}, \frac{3}{30}, \frac{6}{30}, \frac{9}{30}, \frac{11}{30}, \frac{12}{30}, \frac{13}{30}, \frac{14}{30}$

Any set of 10 rational numbers $\frac{k}{30}$ where $-12 < k < 15$ and $k$ is an integer, is a valid answer.

Word Problems Related to Rational Numbers

In previous sections, we learned how to perform basic operations (addition, subtraction, multiplication, and division) with rational numbers. Now, let's apply these operations to solve real-world problems presented in the form of word problems. Solving word problems involves translating the problem description into mathematical expressions or equations using rational numbers and then performing the necessary calculations.

Example 1. A sum of $\textsf{₹}500$ is to be distributed among three people, A, B, and C. A gets $\frac{1}{2}$ of the total sum, B gets $\frac{1}{3}$ of the total sum, and C gets the remaining part of the sum. How much money does C get?

Answer:

Given:

Total sum of money = $\textsf{₹}500$.
A gets $\frac{1}{2}$ of the total sum.
B gets $\frac{1}{3}$ of the total sum.
C gets the rest of the sum.

To Find:

Amount of money C gets.

Solution:

First, calculate the amount A gets.

Amount A gets $= \frac{1}{2} \times \textsf{₹}500$

$= \textsf{₹}\frac{1 \times 500}{2}$

$= \textsf{₹}\frac{500}{2}$

$= \textsf{₹}250$

Next, calculate the amount B gets.

Amount B gets $= \frac{1}{3} \times \textsf{₹}500$

$= \textsf{₹}\frac{1 \times 500}{3}$

$= \textsf{₹}\frac{500}{3}$

Now, find the total amount A and B get together. This involves adding the amounts they received.

Total amount A and B get $= \textsf{₹}250 + \textsf{₹}\frac{500}{3}$

To add an integer and a rational number, treat the integer as a rational number with denominator 1 ($\textsf{₹}250 = \frac{250}{1}$). Find the LCM of the denominators 1 and 3, which is 3.

$\frac{250}{1} + \frac{500}{3} = \frac{250 \times 3}{1 \times 3} + \frac{500}{3} = \frac{750}{3} + \frac{500}{3}$

$= \frac{750+500}{3} = \frac{1250}{3}$

So, A and B together get $\textsf{₹}\frac{1250}{3}$.

Finally, C gets the remaining amount. This is found by subtracting the total amount A and B got from the total sum.

Amount C gets $= \textsf{₹}500 - \textsf{₹}\frac{1250}{3}$

Rewrite 500 as a rational number with denominator 3: $\textsf{₹}500 = \frac{500}{1} = \frac{500 \times 3}{1 \times 3} = \frac{1500}{3}$.

Amount C gets $= \textsf{₹}\frac{1500}{3} - \textsf{₹}\frac{1250}{3}$

$= \textsf{₹}\frac{1500-1250}{3}$

$= \textsf{₹}\frac{250}{3}$

The amount C gets is $\textsf{₹}\frac{250}{3}$. This is a rational number expressed as an improper fraction, which is a valid way to represent the amount. If needed, you could convert it to a mixed number ($83 \frac{1}{3}$) or a decimal ($\approx 83.33$).

Example 2. The product of two rational numbers is $\frac{-12}{35}$. If one of the numbers is $\frac{2}{5}$, find the other number.

Answer:

Given:

Product of two rational numbers $= \frac{-12}{35}$.
One of the rational numbers $= \frac{2}{5}$.

To Find:

The other rational number.

Solution:

Let the other rational number be $x$.

According to the problem, the product of the two numbers is $\frac{-12}{35}$. So, we can write an equation:

$(\text{One number}) \times (\text{Other number}) = \text{Product}$

$\frac{2}{5} \times x = \frac{-12}{35}$

... (i)

To find $x$, we need to isolate it. We can do this by dividing both sides of the equation by the number multiplied with $x$, which is $\frac{2}{5}$. Remember that dividing by a rational number is the same as multiplying by its reciprocal.

The reciprocal (multiplicative inverse) of $\frac{2}{5}$ is $\frac{5}{2}$.

Multiply both sides of equation (i) by $\frac{5}{2}$:

$(\frac{5}{2}) \times (\frac{2}{5} \times x) = (\frac{5}{2}) \times (\frac{-12}{35})$

On the left side, by the associative property of multiplication and the property of multiplicative inverse ($(\frac{5}{2} \times \frac{2}{5}) = 1$):

$(\frac{5}{2} \times \frac{2}{5}) \times x = 1 \times x = x$

On the right side, perform the multiplication $\frac{5}{2} \times \frac{-12}{35}$. We can cancel common factors before multiplying.

$x = \frac{5 \times (-12)}{2 \times 35}$

Cancel 5 in the numerator with 35 in the denominator ($35 \div 5 = 7$). Cancel 2 in the denominator with 12 in the numerator ($12 \div 2 = 6$).

$x = \frac{\cancel{5}^{\normalsize 1} \times (-\cancel{12}^{\normalsize 6})}{\cancel{2}_{\normalsize 1} \times \cancel{35}_{\normalsize 7}}$

Now, multiply the remaining terms:

$x = \frac{1 \times (-6)}{1 \times 7} = \frac{-6}{7}$

The other rational number is $\frac{-6}{7}$.

We can check our answer by multiplying the two numbers:

$\frac{2}{5} \times \frac{-6}{7} = \frac{2 \times (-6)}{5 \times 7} = \frac{-12}{35}$

This matches the given product, so our answer is correct.

Example 3. The perimeter of a rectangle is $13 \text{ cm}$. If its width is $2\frac{3}{4} \text{ cm}$, find its length.

Answer:

Given:

Perimeter of the rectangle = $13 \text{ cm}$.
Width of the rectangle, $w = 2\frac{3}{4} \text{ cm}$.

To Find:

The length of the rectangle, $l$.

Solution:

First, convert the width from a mixed number to an improper fraction:

$w = 2\frac{3}{4} = \frac{(2 \times 4) + 3}{4} = \frac{8+3}{4} = \frac{11}{4} \text{ cm}$

The formula for the perimeter of a rectangle is:

$\text{Perimeter} = 2 \times (\text{length} + \text{width})$

$\text{Perimeter} = 2 \times (l + w)$

Substitute the given values into the formula:

$13 = 2 \times (l + \frac{11}{4})$

To solve for $l$, we can first divide both sides of the equation by 2:

$\frac{13}{2} = l + \frac{11}{4}$

Now, subtract $\frac{11}{4}$ from both sides of the equation to isolate $l$:

$l = \frac{13}{2} - \frac{11}{4}$

To subtract these rational numbers, find the LCM of the denominators 2 and 4. LCM(2, 4) = 4.

Rewrite $\frac{13}{2}$ with denominator 4:

$\frac{13}{2} = \frac{13 \times 2}{2 \times 2} = \frac{26}{4}$

Now perform the subtraction:

$l = \frac{26}{4} - \frac{11}{4}$

$l = \frac{26-11}{4}$

$l = \frac{15}{4}$

The length of the rectangle is $\frac{15}{4} \text{ cm}$. This can be written as a mixed number $3\frac{3}{4} \text{ cm}$.