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Class 7th Chapters
1. Integers 2. Fractions and Decimals 3. Data Handling
4. Simple Equations 5. Lines and Angles 6. The Triangle and its Properties
7. Congruence of Triangles 8. Comparing Quantities 9. Rational Numbers
10. Practical Geometry 11. Perimeter and Area 12. Algebraic Expressions
13. Exponents and Powers 14. Symmetry 15. Visualising Solid Shapes

Content On This Page
Introduction to Rational Numbers Representation of Rational Numbers on Number Line Absolute Value and Comparison of Rational Numbers
Rational Numbers Between Two Rational Numbers Basic Algebraic Operations on Rational Numbers


Chapter 9 Rational Numbers (Concepts)

This chapter significantly expands our understanding of the number system, venturing beyond integers and simple fractions to formally introduce the comprehensive set of Rational Numbers. Building upon prior knowledge, we define a rational number with mathematical precision: it is any number that can be expressed in the specific form $\frac{p}{q}$, where both $p$ (the numerator) and $q$ (the denominator) are integers (whole numbers, their negatives, and zero), with the crucial constraint that the denominator $q$ must not be equal to zero ($q \neq 0$). This definition is remarkably inclusive.

Consider its scope:

Rational numbers can be classified as positive (if the numerator $p$ and denominator $q$ share the same sign, like $\frac{3}{5}$ or $\frac{-2}{-7}$) or negative (if $p$ and $q$ have opposite signs, like $\frac{-3}{5}$ or $\frac{2}{-7}$).

A key practical skill developed here is the accurate representation of rational numbers on the number line. This involves carefully dividing the segments between consecutive integers into a number of equal parts corresponding to the denominator $q$, and then locating the position indicated by the numerator $p$. We also introduce the concept of the standard form (or simplest form) of a rational number. A rational number $\frac{p}{q}$ is in standard form if its denominator $q$ is positive, and the integers $p$ and $q$ have no common factors other than $1$ (they are co-prime). For instance, $\frac{-8}{12}$ is equivalent to $\frac{-2}{3}$ in standard form. Finding equivalent rational numbers is analogous to finding equivalent fractions – achieved by multiplying or dividing both the numerator $p$ and the denominator $q$ by the same non-zero integer $k$ ($\frac{p}{q} = \frac{p \times k}{q \times k}$, where $k \neq 0$).

Comparing the magnitudes of two rational numbers typically involves converting them into equivalent forms that share a common positive denominator (often the Least Common Multiple, LCM, of the original denominators). Once they share the same denominator, the comparison simply reduces to comparing their numerators. A fascinating aspect explored is the density property of rational numbers: between any two distinct rational numbers, no matter how close they appear, there exist infinitely many other rational numbers. We will learn methods to find such numbers, including calculating their average ($\frac{1}{2}(\frac{a}{b} + \frac{c}{d})$) or generating equivalent fractions with progressively larger denominators.

Finally, the chapter systematically defines the four fundamental arithmetic operations for rational numbers, extending the familiar rules for fractions while paying close attention to the sign rules governing integer operations:

The chapter also reinforces essential algebraic properties as they apply to rational numbers under these operations, including closure, commutativity, associativity, the existence of identity elements ($0$ for addition, $1$ for multiplication), and the existence of inverses (additive inverse $-\frac{p}{q}$ for every $\frac{p}{q}$, and multiplicative inverse $\frac{q}{p}$ for every non-zero rational number $\frac{p}{q}$).



Introduction to Rational Numbers

In our journey through mathematics, we started with counting numbers (natural numbers), then included zero to form whole numbers, and later expanded to include negative numbers (integers). We also learned about fractions, which represent parts of a whole. Now, we will bring all these numbers together and introduce a larger collection of numbers called Rational Numbers. Rational numbers encompass all the number types we have studied so far and extend the concept of fractions to include both positive and negative values in the numerator and denominator (with the denominator being non-zero).


What are Rational Numbers?

A Rational Number is formally defined as any number that can be expressed or written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, and $q$ is not equal to zero ($q \neq 0$).

The set of rational numbers is represented by the symbol $\mathbb{Q}$. The symbol $\mathbb{Q}$ comes from the word "quotient," as a rational number is essentially a quotient of two integers.

Examples of Rational Numbers:

Let's see how different types of numbers fit into the definition of rational numbers:

Numbers that cannot be expressed in the form $\frac{p}{q}$ (where $p, q$ are integers and $q \neq 0$) are called irrational numbers (like $\sqrt{2}$ or $\pi$). You will learn about these in higher classes.

Numerator and Denominator of a Rational Number:

In a rational number written as $\frac{p}{q}$ (where $p$ and $q$ are integers and $q \neq 0$):


Positive and Negative Rational Numbers

Similar to integers, rational numbers can be positive, negative, or zero.

1. Positive Rational Numbers

A rational number is considered Positive if its numerator and its denominator have the same sign. This means either both are positive, or both are negative.

Examples:

All positive fractions and positive integers are positive rational numbers.

2. Negative Rational Numbers

A rational number is considered Negative if its numerator and its denominator have opposite signs. This means one is positive and the other is negative.

Examples:

All negative fractions and negative integers are negative rational numbers.

3. Zero

The rational number zero ($0$) is neither a positive rational number nor a negative rational number. It can be expressed as $\frac{0}{q}$ where $q$ is any non-zero integer (e.g., $\frac{0}{5}$, $\frac{0}{-10}$). The numerator is 0, but the denominator is non-zero, satisfying the definition of a rational number.


Standard Form of a Rational Number

A rational number $\frac{p}{q}$ is said to be in its Standard Form (or lowest terms or simplest form) if it satisfies two conditions:

  1. The denominator $q$ is a positive integer.
  2. The numerator $p$ and the denominator $q$ have no common factor other than 1 (i.e., the HCF of $|p|$ and $|q|$ is 1).

To convert any given rational number into its standard form, follow these steps:

  1. If the denominator is negative, make it positive by multiplying both the numerator and the denominator by -1.
  2. Find the Highest Common Factor (HCF) of the absolute values of the numerator and the (now positive) denominator.
  3. Divide both the numerator and the denominator by their HCF. The resulting fraction will be in standard form.

Example: Express $\frac{-12}{30}$ in standard form.

Step 1: The denominator (30) is already positive.

Step 2: The numerator is -12 and the denominator is 30. Find the HCF of their absolute values, $|-12|=12$ and $|30|=30$.

Factors of 12: 1, 2, 3, 4, 6, 12.

Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.

The common factors are 1, 2, 3, 6. The HCF is 6.

Step 3: Divide both the numerator and the denominator by their HCF (6).

$\frac{-12 \div 6}{30 \div 6} = \frac{-2}{5} $.

The standard form of $\frac{-12}{30}$ is $\frac{-2}{5}$. The denominator (5) is positive, and the HCF of |-2| and 5 is 1.

Example: Express $\frac{14}{-49}$ in standard form.

Step 1: The denominator (-49) is negative. Make it positive by multiplying numerator and denominator by -1.

$\frac{14}{-49} = \frac{14 \times (-1)}{-49 \times (-1)} = \frac{-14}{49} $.

Step 2: The numerator is -14, and the denominator is 49. Find the HCF of their absolute values, $|-14|=14$ and $|49|=49$.

Factors of 14: 1, 2, 7, 14.

Factors of 49: 1, 7, 49.

The HCF is 7.

Step 3: Divide both the numerator and the denominator by their HCF (7).

$\frac{-14 \div 7}{49 \div 7} = \frac{-2}{7} $.

The standard form of $\frac{14}{-49}$ is $\frac{-2}{7}$. The denominator (7) is positive, and HCF of |-2| and 7 is 1.


Equivalent Rational Numbers

Equivalent Rational Numbers are rational numbers that represent the same value. Just like equivalent fractions, they can be obtained by multiplying or dividing both the numerator and the denominator of a given rational number by the same non-zero integer.

If $\frac{p}{q}$ is a rational number, and $m$ is any non-zero integer, then $\frac{p \times m}{q \times m}$ is an equivalent rational number to $\frac{p}{q}$.

If $\frac{p}{q}$ is a rational number, and $k$ is a common divisor of $p$ and $q$ (and $k \neq 0$), then $\frac{p \div k}{q \div k}$ is an equivalent rational number to $\frac{p}{q}$.

Example: Write three rational numbers equivalent to $\frac{-2}{7}$.

Multiply numerator and denominator by different non-zero integers:

1. Multiply by 2: $\frac{-2 \times 2}{7 \times 2} = \frac{-4}{14}$.

2. Multiply by 3: $\frac{-2 \times 3}{7 \times 3} = \frac{-6}{21}$.

3. Multiply by -1: $\frac{-2 \times (-1)}{7 \times (-1)} = \frac{2}{-7}$.

So, $\frac{-4}{14}, \frac{-6}{21}, \frac{2}{-7}$ are three rational numbers equivalent to $\frac{-2}{7}$. (There are infinitely many such numbers).

Checking for Equivalence:

Two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equivalent if and only if their fractional forms are equal, i.e., $\frac{a}{b} = \frac{c}{d}$. This is true if the product of the numerator of the first and the denominator of the second is equal to the product of the denominator of the first and the numerator of the second (cross-product rule): $a \times d = b \times c$.

Example: Are $\frac{-3}{5}$ and $\frac{9}{-15}$ equivalent rational numbers?

Let $\frac{a}{b} = \frac{-3}{5}$ (so $a=-3, b=5$) and $\frac{c}{d} = \frac{9}{-15}$ (so $c=9, d=-15$).

Check if $a \times d = b \times c$.

Calculate $a \times d$: $(-3) \times (-15)$. Multiply absolute values $3 \times 15 = 45$. Product of two negatives is positive. So, $(-3) \times (-15) = 45$.

Calculate $b \times c$: $5 \times 9 = 45$.

Compare the products: $45$ and $45$.

Since $a \times d = b \times c$ ($45 = 45$), the rational numbers $\frac{-3}{5}$ and $\frac{9}{-15}$ are equivalent.

(Alternatively, we could simplify $\frac{9}{-15}$ to its standard form: $\frac{9}{-15} = \frac{9 \times (-1)}{-15 \times (-1)} = \frac{-9}{15}$. HCF(9, 15) = 3. $\frac{-9 \div 3}{15 \div 3} = \frac{-3}{5}$. Since the standard form is the same as the first number, they are equivalent).

Example 1. List five rational numbers between -1 and 0.

Answer:

We need to find five rational numbers that lie between -1 and 0 on the number line. We can write -1 and 0 as rational numbers with a common denominator to create 'space' between them.

Write -1 as a rational number: $-1 = \frac{-1}{1}$.

Write 0 as a rational number: $0 = \frac{0}{1}$.

To find numbers between $\frac{-1}{1}$ and $\frac{0}{1}$, we can write equivalent rational numbers by multiplying the numerator and denominator by the same number. Since we need 5 numbers, let's choose a denominator larger than $5+1=6$, say 10. Multiply the numerator and denominator of both numbers by 10:

$-1 = \frac{-1}{1} = \frac{-1 \times 10}{1 \times 10} = \frac{-10}{10} $.

$0 = \frac{0}{1} = \frac{0 \times 10}{1 \times 10} = \frac{0}{10} $.

Now we need to find rational numbers between $\frac{-10}{10}$ and $\frac{0}{10}$. The numerators are -10 and 0, and the denominator is 10. We can pick any integers between -10 and 0 and use them as numerators with the denominator 10.

Integers between -10 and 0 are -9, -8, -7, -6, -5, -4, -3, -2, -1.

We can choose any five of these integers as numerators. For example, -9, -8, -7, -6, -5.

Five rational numbers between $\frac{-10}{10}$ and $\frac{0}{10}$ are:

$\frac{-9}{10}, \frac{-8}{10}, \frac{-7}{10}, \frac{-6}{10}, \frac{-5}{10} $.

These numbers are between -1 and 0. We should write them in their simplest form if possible:

  • $\frac{-9}{10}$ (Standard form)
  • $\frac{-8}{10} = \frac{-4}{5}$ (Standard form)
  • $\frac{-7}{10}$ (Standard form)
  • $\frac{-6}{10} = \frac{-3}{5}$ (Standard form)
  • $\frac{-5}{10} = \frac{-1}{2}$ (Standard form)

Thus, five rational numbers between -1 and 0 are $\frac{-9}{10}, \frac{-4}{5}, \frac{-7}{10}, \frac{-3}{5}, \frac{-1}{2}$.

(Note: Many other sets of 5 numbers are possible, depending on the common denominator chosen and which integers are selected as numerators).

Example 2. Write the following rational numbers in standard form:

(a) $\frac{-45}{30}$

(b) $\frac{36}{-24}$

Answer:

To write a rational number in standard form, we ensure the denominator is positive and the numerator and denominator have no common factors other than 1.

(a) Express $\frac{-45}{30}$ in standard form.

Step 1: Check the denominator. The denominator (30) is positive. This condition is met.

Step 2: Find the HCF of the absolute values of the numerator and the denominator. Absolute value of numerator $|-45| = 45$. Denominator is 30. Find HCF(45, 30).

Factors of 45: 1, 3, 5, 9, 15, 45.

Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.

The common factors are 1, 3, 5, 15. The Highest Common Factor (HCF) is 15.

Step 3: Divide both the numerator and the denominator by their HCF (15).

$\frac{-45 \div 15}{30 \div 15} = \frac{-3}{2} $.

The standard form of $\frac{-45}{30}$ is $\frac{-3}{2}$.

(b) Express $\frac{36}{-24}$ in standard form.

Step 1: Check the denominator. The denominator (-24) is negative. Make it positive by multiplying both the numerator and the denominator by -1.

$\frac{36}{-24} = \frac{36 \times (-1)}{-24 \times (-1)} = \frac{-36}{24} $.

Step 2: Find the HCF of the absolute values of the numerator and the (now positive) denominator. Absolute value of numerator $|-36| = 36$. Denominator is 24. Find HCF(36, 24).

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24.

The common factors are 1, 2, 3, 4, 6, 12. The Highest Common Factor (HCF) is 12.

Step 3: Divide both the numerator and the denominator by their HCF (12).

$\frac{-36 \div 12}{24 \div 12} = \frac{-3}{2} $.

The standard form of $\frac{36}{-24}$ is $\frac{-3}{2}$.



Representation of Rational Numbers on Number Line

We have learned to represent whole numbers, integers, and fractions on a number line. Rational numbers are a broader set that includes all these types. Just as we can visualise integers and fractions on a number line, we can also represent any rational number as a unique point on the number line. This graphical representation helps us understand the order and relative values of rational numbers, including positive and negative fractions.


The Number Line for Rational Numbers

The number line is a straight line with a point designated as the origin (representing zero). Positive numbers are located at equal intervals to the right of zero, and negative numbers are located at the same equal intervals to the left of zero. The number line extends infinitely in both positive and negative directions.

A number line with zero at the center, positive integers (1, 2, 3...) to the right, and negative integers (-1, -2, -3...) to the left. Points between integers represent fractions/decimals.

Integers are marked at specific points. Fractions and decimals lie between these integer points. Rational numbers fill up the number line, although there are points on the number line that are not rational (these represent irrational numbers, like $\sqrt{2}$ or $\pi$, which you will encounter later).


Steps to Represent a Rational Number $\frac{p}{q}$ on the Number Line

Here, $p$ and $q$ are integers, and $q \neq 0$. Assume $q > 0$ (if $q$ is negative, convert $\frac{p}{q}$ to an equivalent rational number with a positive denominator first, e.g., $\frac{3}{-4} = \frac{-3}{4}$).

To represent a rational number $\frac{p}{q}$ (where $q > 0$) on the number line:

  1. Determine the Sign:
    • If the rational number is positive (i.e., $p > 0$), it will lie to the right of 0 on the number line.
    • If the rational number is negative (i.e., $p < 0$), it will lie to the left of 0 on the number line.
    • If $p = 0$, the rational number is 0, located at the origin.
  2. Identify the Integers it Lies Between:
    • If the absolute value of the rational number is less than 1 (i.e., $|p| < |q|$, it's a proper fraction or its negative), it lies between 0 and 1 (if positive) or between -1 and 0 (if negative).
    • If the absolute value of the rational number is greater than or equal to 1 (i.e., $|p| \ge |q|$, it's an improper fraction or its negative), convert it into a mixed number first. For a positive rational number $W\frac{P}{D}$, it lies between the integer $W$ and $W+1$. For a negative rational number $-W\frac{P}{D}$, it lies between the integer $-W$ and $-(W+1)$ (i.e., between $-W$ and $-W-1$, or between $-W-1$ and $-W$ if listed in increasing order).
    • Example: $\frac{7}{4} = 1\frac{3}{4}$. This is between 1 and 2.

      Example: $\frac{-5}{3} = -1\frac{2}{3}$. This is between -2 and -1.

  3. Divide the Unit Segment:

    Identify the unit segment on the number line between the two integers determined in Step 2 (e.g., [0, 1], [1, 2], [-1, 0], [-2, -1]).

    Divide this unit segment into $q$ (the positive denominator) equal parts. The number of division marks you make will be $q-1$.

  4. Locate the Point:

    Starting from the left integer of the segment (for positive rational numbers) or from the right integer of the segment (for negative rational numbers when moving left from zero), count $|p|$ of these equal parts.

    Mark the point at the end of this count. This point represents the rational number $\frac{p}{q}$.


Examples of Representation

Example 1: Represent $\frac{3}{4}$ on the number line.

1. Sign: $\frac{3}{4}$ is positive ($p=3, q=4$, both positive). It lies to the right of 0.

2. Integers: $|\frac{3}{4}| = \frac{3}{4} < 1$. It's a proper fraction, so it lies between 0 and 1.

3. Divide: The denominator is 4 ($q=4$). Divide the segment between 0 and 1 into 4 equal parts. (Make 3 division marks).

Number line segment between 0 and 1 divided into 4 equal parts.

The points of division from left to right are $\frac{1}{4}, \frac{2}{4}, \frac{3}{4}$ (relative to 0, where $0=\frac{0}{4}$ and $1=\frac{4}{4}$).

4. Locate: The numerator is 3 ($p=3$). Starting from 0, count 3 parts to the right.

Number line showing the point 3/4 located by counting 3 parts from 0 in the segment 0-1 which is divided into 4 parts.

The point P represents $\frac{3}{4}$.

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

1. Sign: $\frac{-2}{3}$ is negative ($p=-2, q=3$). It lies to the left of 0.

2. Integers: $|\frac{-2}{3}| = \frac{2}{3} < 1$. It's a proper fraction (negative), so it lies between -1 and 0.

3. Divide: The positive denominator is 3 ($q=3$). Divide the segment between -1 and 0 into 3 equal parts. (Make 2 division marks).

Number line segment between 0 and -1 divided into 3 equal parts.

Moving left from 0, the division points represent $\frac{-1}{3}, \frac{-2}{3}$ (relative to 0, where $0=\frac{0}{3}$ and $-1=\frac{-3}{3}$).

4. Locate: The absolute value of the numerator is 2 ($|p|=2$). Starting from 0, count 2 parts to the left.

Number line showing the point -2/3 located by counting 2 parts from 0 to the left in the segment 0 to -1 which is divided into 3 parts.

The point Q represents $\frac{-2}{3}$.

Example 3: Represent $\frac{7}{5}$ on the number line.

1. Sign: $\frac{7}{5}$ is positive ($p=7, q=5$). It lies to the right of 0.

2. Integers: $|\frac{7}{5}| = \frac{7}{5} > 1$. It's an improper fraction. Convert to mixed number: $\frac{7}{5} = 1\frac{2}{5}$. This tells us $\frac{7}{5}$ lies between the integers 1 and 2. Specifically, it is 1 whole unit plus $\frac{2}{5}$ of the next unit.

3. Divide: The fractional part is $\frac{2}{5}$. The denominator is 5 ($q=5$). Divide the unit segment between 1 and 2 into 5 equal parts. (Make 4 division marks).

Number line segment between 1 and 2 divided into 5 equal parts.

The division points from 1 to 2 represent $1\frac{1}{5}, 1\frac{2}{5}, 1\frac{3}{5}, 1\frac{4}{5}$ (or $\frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}$, where $1 = \frac{5}{5}$ and $2 = \frac{10}{5}$).

4. Locate: The numerator of the fractional part is 2. Starting from 1, count 2 parts to the right.

Number line showing the point 7/5 (or 1 and 2/5) located by counting 2 parts from 1 in the segment 1-2 which is divided into 5 parts.

The point R represents $\frac{7}{5}$ (or $1\frac{2}{5}$).

Example 4: Represent $\frac{-11}{4}$ on the number line.

1. Sign: $\frac{-11}{4}$ is negative ($p=-11, q=4$). It lies to the left of 0.

2. Integers: $|\frac{-11}{4}| = \frac{11}{4}$. Convert to mixed number: $\frac{11}{4} = 2\frac{3}{4}$. So, $\frac{-11}{4} = -2\frac{3}{4}$. This means the number is 2 whole units to the left of 0, plus another $\frac{3}{4}$ of the next unit. It lies between the integers -2 and -3.

3. Divide: The fractional part is $\frac{3}{4}$. The denominator is 4 ($q=4$). Divide the unit segment between -2 and -3 into 4 equal parts. (Make 3 division marks).

Number line segment between -2 and -3 divided into 4 equal parts.

Moving left from -2, the division points represent $-2\frac{1}{4}, -2\frac{2}{4}, -2\frac{3}{4}$ (or $\frac{-9}{4}, \frac{-10}{4}, \frac{-11}{4}$, where $-2 = \frac{-8}{4}$ and $-3 = \frac{-12}{4}$).

4. Locate: The numerator of the fractional part is 3. Starting from -2, count 3 parts to the left.

Number line showing the point -11/4 (or -2 and 3/4) located by counting 3 parts from -2 to the left in the segment -2 to -3 which is divided into 4 parts.

The point S represents $\frac{-11}{4}$ (or $-2\frac{3}{4}$).


Key Observation: Every rational number corresponds to a unique point on the number line. You can find a point on the number line for any rational number $\frac{p}{q}$. However, there are points on the number line that do not represent rational numbers (e.g., the points corresponding to $\sqrt{2}$ or $\pi$). The collection of all rational numbers and all irrational numbers together form the complete real number line.

Example 1. Represent the rational number $\frac{5}{8}$ on the number line.

Answer:

We need to represent the rational number $\frac{5}{8}$ on the number line.

1. Sign: $\frac{5}{8}$ is positive ($5 > 0, 8 > 0$). It will be located to the right of 0.

2. Integers it Lies Between: The absolute value $|\frac{5}{8}| = \frac{5}{8}$. Since $5 < 8$, this is a proper fraction, and it is positive. So, $\frac{5}{8}$ lies between the integers 0 and 1.

3. Divide the Unit Segment: The denominator is 8. We need to divide the segment of the number line between 0 and 1 into 8 equal parts.

This requires making $8-1 = 7$ equally spaced division marks between 0 and 1. These points will correspond to $\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}$.

Number line from 0 to 1, divided into 8 equal parts.

4. Locate the Point: The numerator is 5. Starting from 0, we move 5 of these equal parts to the right.

Number line from 0 to 1, divided into 8 equal parts. The 5th mark from 0 is labeled as 5/8.

The point representing $\frac{5}{8}$ is the fifth mark to the right of 0 within the segment [0, 1] that has been divided into 8 equal parts.



Absolute Value and Comparison of Rational Numbers

In the previous section, we introduced rational numbers as an extension of integers and fractions. Just like integers, rational numbers can be positive or negative. To work with them effectively, particularly for operations and comparisons, it's important to understand the concept of their absolute value and how to compare their magnitudes and positions on the number line.


Absolute Value of a Rational Number

The Absolute Value of a rational number is its distance from zero on the number line, irrespective of its direction (left or right of zero). Since distance is always a non-negative quantity, the absolute value of any rational number is always non-negative (zero or positive).

The absolute value of a rational number $x$ is denoted by writing the number within two vertical bars, $|x|$.

If $x$ is a rational number:

For a rational number $x = \frac{p}{q}$, the absolute value can be calculated as the absolute value of the numerator divided by the absolute value of the denominator:

$|\frac{p}{q}| = \frac{|p|}{|q|} $.

Since we usually write rational numbers in standard form with a positive denominator ($q>0$), this often simplifies to $\frac{|p|}{q}$.

Examples of Absolute Value:

Key Points:


Comparison of Rational Numbers

Comparing rational numbers means determining which of two or more rational numbers is greater than, less than, or equal to the others. This is equivalent to determining their relative positions on the number line: the number further to the right is always greater.

1. Comparing Positive Rational Numbers

Comparing positive rational numbers is the same as comparing positive fractions, which you learned in Class 6. To compare two positive rational numbers, say $\frac{a}{b}$ and $\frac{c}{d}$ (where $a, b, c, d$ are positive integers):

Method 1: Convert to Like Fractions (Common Denominator)

  1. Find the Least Common Multiple (LCM) of the denominators $b$ and $d$.
  2. Convert both rational numbers into equivalent rational numbers with this LCM as their common denominator.
  3. Compare the numerators of these new equivalent fractions. The rational number with the greater numerator is the greater rational number.

Example: Compare $\frac{2}{3}$ and $\frac{4}{5}$.

LCM of denominators 3 and 5 is 15.

Convert to equivalent fractions with denominator 15:

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} $.

$\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} $.

Now compare the numerators 10 and 12. Since $10 < 12$, we have $\frac{10}{15} < \frac{12}{15}$.

Therefore, $\frac{2}{3} < \frac{4}{5}$.

Method 2: Cross-Multiplication

To compare two positive rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ (with $a, b, c, d > 0$), compare the cross-products $a \times d$ and $b \times c$.

Example: Compare $\frac{2}{3}$ and $\frac{4}{5}$ using cross-multiplication.

Here, $a=2, b=3, c=4, d=5$.

Calculate $a \times d = 2 \times 5 = 10$.

Calculate $b \times c = 3 \times 4 = 12$.

Compare the cross-products: $10$ and $12$. Since $10 < 12$, $\frac{a}{b} < \frac{c}{d}$.

Therefore, $\frac{2}{3} < \frac{4}{5}$.

2. Comparing Negative Rational Numbers

To compare two negative rational numbers:

  1. First, it is helpful to write them with positive denominators if they are not already in that form. For example, $\frac{2}{-7} = \frac{-2}{7}$.
  2. Then, compare their absolute values using the methods for positive rational numbers.
  3. The negative rational number with the smaller absolute value is the greater number (because it is located closer to zero on the left side of the number line).

Example: Compare $\frac{-2}{3}$ and $\frac{-4}{5}$.

Step 1: Both have positive denominators.

Step 2: Compare their absolute values: $|\frac{-2}{3}| = \frac{2}{3}$ and $|\frac{-4}{5}| = \frac{4}{5}$.

From the previous example, we found that $\frac{2}{3} < \frac{4}{5}$. So, $|\frac{-2}{3}| < |\frac{-4}{5}|$.

Step 3: Since $\frac{-2}{3}$ has a smaller absolute value than $\frac{-4}{5}$, $\frac{-2}{3}$ is closer to 0 on the number line.

On the number line, numbers to the right are greater. Since $\frac{-2}{3}$ is to the right of $\frac{-4}{5}$ (because $-0.66... > -0.8$), $\frac{-2}{3}$ is greater.

Therefore, $\frac{-2}{3} > \frac{-4}{5}$.

3. Comparing a Positive and a Negative Rational Number

This is straightforward based on their positions on the number line. Any positive rational number is always located to the right of zero, and any negative rational number is always located to the left of zero. Therefore, any positive rational number is always greater than any negative rational number.

Example: Compare $\frac{1}{2}$ and $\frac{-3}{4}$.

Since $\frac{1}{2}$ is a positive rational number and $\frac{-3}{4}$ is a negative rational number, $\frac{1}{2}$ is greater than $\frac{-3}{4}$.

$\frac{1}{2} > \frac{-3}{4} $.

4. Comparing with Zero


Example 1. Find the absolute value of: (a) $\frac{-7}{11}$ (b) $\frac{5}{9}$ (c) $\frac{0}{-4}$

Answer:

(a) Find the absolute value of $\frac{-7}{11}$.

The number $\frac{-7}{11}$ is negative. Its absolute value is its positive counterpart, or using the formula $|x| = -x$ for $x < 0$.

$|\frac{-7}{11}| = -(\frac{-7}{11}) = \frac{7}{11} $.

Alternatively, using $| \frac{p}{q} | = \frac{|p|}{|q|}$, $|\frac{-7}{11}| = \frac{|-7|}{|11|} = \frac{7}{11}$.

(b) Find the absolute value of $\frac{5}{9}$.

The number $\frac{5}{9}$ is positive. Its absolute value is the number itself.

$|\frac{5}{9}| = \frac{5}{9} $.

(c) Find the absolute value of $\frac{0}{-4}$.

First, simplify the rational number: $\frac{0}{-4} = 0$.

The absolute value of 0 is 0.

$|\frac{0}{-4}| = |0| = 0 $.

Example 2. Which is greater: $\frac{-5}{6}$ or $\frac{-4}{3}$?

Answer:

We need to compare two negative rational numbers: $\frac{-5}{6}$ and $\frac{-4}{3}$.

Method 1: Convert to like fractions (with positive denominators) and compare numerators.

The denominators are 6 and 3. Find the LCM of 6 and 3. LCM(6, 3) = 6.

Convert both rational numbers to equivalent forms with denominator 6:

  • $\frac{-5}{6}$ is already in this form.
  • $\frac{-4}{3} = \frac{-4 \times 2}{3 \times 2} = \frac{-8}{6}$.

Now we compare $\frac{-5}{6}$ and $\frac{-8}{6}$. Since the denominators are the same and positive, we compare the numerators: -5 and -8.

On the number line, -5 is to the right of -8, so $-5 > -8$.

Therefore, $\frac{-5}{6} > \frac{-8}{6}$. Since $\frac{-8}{6}$ is equivalent to $\frac{-4}{3}$, we have $\frac{-5}{6} > \frac{-4}{3}$.

Method 2: Compare absolute values.

Find the absolute value of each number:

$|\frac{-5}{6}| = \frac{5}{6} $.

$|\frac{-4}{3}| = \frac{4}{3} $.

Now compare the positive rational numbers $\frac{5}{6}$ and $\frac{4}{3}$. LCM(6, 3) = 6.

Convert $\frac{4}{3}$ to denominator 6: $\frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6}$.

Compare $\frac{5}{6}$ and $\frac{8}{6}$. Since $5 < 8$, $\frac{5}{6} < \frac{8}{6}$. So, $|\frac{-5}{6}| < |\frac{-4}{3}|$.

For negative numbers, the one with the smaller absolute value is greater.

Therefore, $\frac{-5}{6} > \frac{-4}{3}$.

Both methods give the same result. $\frac{-5}{6}$ is greater than $\frac{-4}{3}$.

Example 3. Arrange the rational numbers $\frac{-3}{5}, \frac{7}{-10}, \frac{-5}{8}$ in ascending order.

Answer:

The given rational numbers are $\frac{-3}{5}, \frac{7}{-10}, \frac{-5}{8}$.

Step 1: Write each rational number with a positive denominator to make comparison easier.

  • $\frac{-3}{5}$ already has a positive denominator (5).
  • $\frac{7}{-10}$. The denominator (-10) is negative. Multiply numerator and denominator by -1: $\frac{7}{-10} = \frac{7 \times (-1)}{-10 \times (-1)} = \frac{-7}{10}$.
  • $\frac{-5}{8}$ already has a positive denominator (8).

So, we need to compare the rational numbers $\frac{-3}{5}, \frac{-7}{10}, \frac{-5}{8}$. All are negative.

Step 2: Find the LCM of the denominators 5, 10, and 8 to convert them to like fractions.

$ \begin{array}{c|ccc} 2 & 5 \; , & 10 \; , & 8 \\ \hline 2 & 5 \; , & 5 \; , & 4 \\ \hline 2 & 5 \; , & 5 \; , & 2 \\ \hline 5 & 5 \; , & 5 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \end{array} $

LCM of 5, 10, and 8 is $2 \times 2 \times 2 \times 5 = 8 \times 5 = 40$.

Step 3: Convert each rational number to an equivalent rational number with denominator 40.

  • For $\frac{-3}{5}$: $40 \div 5 = 8$. Multiply numerator and denominator by 8: $\frac{-3 \times 8}{5 \times 8} = \frac{-24}{40}$.
  • For $\frac{-7}{10}$: $40 \div 10 = 4$. Multiply numerator and denominator by 4: $\frac{-7 \times 4}{10 \times 4} = \frac{-28}{40}$.
  • For $\frac{-5}{8}$: $40 \div 8 = 5$. Multiply numerator and denominator by 5: $\frac{-5 \times 5}{8 \times 5} = \frac{-25}{40}$.

Step 4: Compare the numerators of these equivalent rational numbers: -24, -28, -25. Remember that for negative numbers, the smaller the number (more negative), the less its value.

Arranging these numerators in ascending order (from smallest to largest):

$-28 < -25 < -24 $.

Step 5: Write the corresponding rational numbers in ascending order.

$\frac{-28}{40} < \frac{-25}{40} < \frac{-24}{40} $.

Step 6: Replace the equivalent fractions with their original forms to get the final answer in terms of the original rational numbers.

  • $\frac{-28}{40}$ corresponds to $\frac{-7}{10}$ (which was $\frac{7}{-10}$).
  • $\frac{-25}{40}$ corresponds to $\frac{-5}{8}$.
  • $\frac{-24}{40}$ corresponds to $\frac{-3}{5}$.

Therefore, the ascending order is $\frac{7}{-10}, \frac{-5}{8}, \frac{-3}{5} $.



Rational Numbers Between Two Rational Numbers

In the world of numbers, we've seen that between any two consecutive integers (like 1 and 2, or -5 and -4), there are no other integers. However, the set of rational numbers has a different property: it is 'dense'. The Density Property of rational numbers states that between any two distinct rational numbers, there are infinitely many other rational numbers.

This means that no matter how close two rational numbers are to each other on the number line, you can always find another rational number that lies somewhere in between them.

Methods to Find Rational Numbers Between Two Given Rational Numbers

Since there are infinitely many rational numbers between any two distinct ones, we can use different methods to find some of them. Here are two common methods:

Method 1: Using the Concept of Average (Midpoint)

The average of two numbers is always located exactly halfway between them. If you take the average of two rational numbers, the result will be a rational number that lies between the original two.

If $a$ and $b$ are two distinct rational numbers, then their average, $\frac{a+b}{2}$, is a rational number that lies between $a$ and $b$.

Rational Number between $a$ and $b$ $= \frac{a+b}{2} $.

You can use this method repeatedly. Once you find a number between $a$ and $b$, you can then find a number between $a$ and this new number, or between the new number and $b$, and so on, generating as many rational numbers as you need.

Example: Find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$ using the average method.

Let $a = \frac{1}{3}$ and $b = \frac{1}{2}$.

The rational number between them is the average of $a$ and $b$: $\frac{a+b}{2} = \frac{\frac{1}{3} + \frac{1}{2}}{2}$.

First, calculate the sum $\frac{1}{3} + \frac{1}{2}$. Find a common denominator for 3 and 2. LCM(3, 2) = 6.

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

Now, divide this sum by 2 (or multiply by $\frac{1}{2}$, the reciprocal of 2):

$\frac{5}{6} \div 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 $\frac{1}{3}$ and $\frac{1}{2}$.

We can verify this: $\frac{1}{3} = \frac{4}{12}$ and $\frac{1}{2} = \frac{6}{12}$. Indeed, $\frac{4}{12} < \frac{5}{12} < \frac{6}{12}$.

To find another number, average $\frac{1}{3}$ and $\frac{5}{12}$:

$\frac{\frac{1}{3} + \frac{5}{12}}{2} = \frac{\frac{4}{12} + \frac{5}{12}}{2} = \frac{\frac{9}{12}}{2} = \frac{9}{12} \times \frac{1}{2} = \frac{9}{24} = \frac{3}{8}$. So, $\frac{1}{3} < \frac{3}{8} < \frac{5}{12} < \frac{1}{2}$.

Method 2: Using Equivalent Rational Numbers with a Common Denominator

This is often a more practical method for finding multiple rational numbers between two given ones. The idea is to rewrite the two rational numbers with a sufficiently large common denominator so that there are integers between their new numerators.

Steps:

  1. Convert the given rational numbers to equivalent rational numbers with the same (positive) denominator. Find the LCM of the original denominators.
  2. If there are enough integers between the new numerators to list the required number of rational numbers, then the numbers of the form $\frac{\text{Integer}}{\text{Common Denominator}}$ between the two become your answer.
  3. If there are not enough integers between the numerators, find new equivalent fractions with an even larger common denominator. You can do this by multiplying the numerators and denominators of the equivalent fractions from Step 1 by the same suitable integer (e.g., 2, 3, 10, or whatever is needed to create enough integer values between the numerators).
  4. List the required number of rational numbers with the common denominator and numerators being the integers between the new numerators. Simplify these rational numbers if required.

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

Step 1: Find a common denominator for 5 and 2. LCM(5, 2) = 10.

Convert the numbers to equivalent rational 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} $.

Step 2: Now we need to find five rational numbers between $\frac{-4}{10}$ and $\frac{5}{10}$. We look for integers between the numerators -4 and 5. These integers are -3, -2, -1, 0, 1, 2, 3, 4.

There are more than five integers between -4 and 5. We can choose any five of these integers as numerators with the denominator 10.

Five rational numbers are: $\frac{-3}{10}, \frac{-2}{10}, \frac{-1}{10}, \frac{0}{10}, \frac{1}{10}$.

Step 3 (Optional): Simplify the fractions if required.

Thus, five rational numbers between $\frac{-2}{5}$ and $\frac{1}{2}$ are $\frac{-3}{10}, \frac{-1}{5}, \frac{-1}{10}, 0, \frac{1}{10}$.

Example (Needing a larger denominator): Find three rational numbers between $\frac{1}{4}$ and $\frac{1}{3}$.

Step 1: Find a common denominator for 4 and 3. LCM(4, 3) = 12.

Convert to equivalent rational numbers with denominator 12:

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

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

Step 2: We need to find three rational numbers between $\frac{3}{12}$ and $\frac{4}{12}$. Look for integers between the numerators 3 and 4. There are no integers between 3 and 4.

Step 3: Since there are not enough integers between the numerators, we need to find equivalent fractions with a larger common denominator. Multiply the numerators and denominators of $\frac{3}{12}$ and $\frac{4}{12}$ by a suitable integer, say 5 (any integer greater than the number of required numbers (3) would work, but 5 provides a good gap). Multiplying by 4 would give a gap of $4 \times 1 = 4$ between $3 \times 4 = 12$ and $4 \times 4 = 16$, giving 3 integers (13, 14, 15). Let's use 5.

$\frac{3}{12} = \frac{3 \times 5}{12 \times 5} = \frac{15}{60} $.

$\frac{4}{12} = \frac{4 \times 5}{12 \times 5} = \frac{20}{60} $.

Now we need to find rational numbers between $\frac{15}{60}$ and $\frac{20}{60}$. The integers between the numerators 15 and 20 are: 16, 17, 18, 19.

Step 4: We need three numbers. We can choose any three of these integers as numerators with the denominator 60.

Three rational numbers are: $\frac{16}{60}, \frac{17}{60}, \frac{18}{60}$.

(Optional) Simplify the fractions: $\frac{16}{60} = \frac{4}{15}$, $\frac{17}{60}$ (simplest form), $\frac{18}{60} = \frac{3}{10}$.

Thus, three rational numbers between $\frac{1}{4}$ and $\frac{1}{3}$ could be $\frac{4}{15}, \frac{17}{60}, \frac{3}{10}$. (Other answers are possible depending on the multiplier used in Step 3 and which integers are chosen).

Example 1. List three rational numbers between -2 and -1.

Answer:

We need to find three rational numbers between the integers -2 and -1. Integers can be written as rational numbers with denominator 1: $-2 = \frac{-2}{1}$ and $-1 = \frac{-1}{1}$.

We need to create a gap between the numerators so we can find 3 integers in between. Multiply the numerator and denominator of both numbers by an integer greater than $3+1=4$, say 5.

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

$-1 = \frac{-1}{1} = \frac{-1 \times 5}{1 \times 5} = \frac{-5}{5} $.

Now we need to find three rational numbers between $\frac{-10}{5}$ and $\frac{-5}{5}$. We look for integers between the numerators -10 and -5. These integers are -9, -8, -7, -6.

We need three numbers, and we have four integers available. We can choose any three of these integers as numerators with the denominator 5.

Three rational numbers between $\frac{-10}{5}$ and $\frac{-5}{5}$ are, for example: $\frac{-9}{5}, \frac{-8}{5}, \frac{-7}{5}$.

(Optional) Simplify the fractions if possible (none of these are in lowest terms, but their numerators and denominators have no common factors other than 1, so they are in standard form).

Thus, three rational numbers between -2 and -1 are $\frac{-9}{5}, \frac{-8}{5}, \frac{-7}{5}$.

(Verification: $-2 = -2.0$, $\frac{-9}{5} = -1.8$, $\frac{-8}{5} = -1.6$, $\frac{-7}{5} = -1.4$, $-1 = -1.0$. These are indeed between -2 and -1 and in order: $-2 < -1.8 < -1.6 < -1.4 < -1$).

Example 2. Find ten rational numbers between $\frac{-5}{6}$ and $\frac{5}{8}$.

Answer:

Given rational numbers are $\frac{-5}{6}$ and $\frac{5}{8}$.

Step 1: Find a common denominator for 6 and 8. LCM(6, 8) = 24.

Convert the given rational numbers to equivalent rational numbers with denominator 24:

$\frac{-5}{6} = \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24} $.

$\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24} $.

Now we need to find ten rational numbers between $\frac{-20}{24}$ and $\frac{15}{24}$. Look for integers between the numerators -20 and 15.

The integers between -20 and 15 are: -19, -18, -17, ..., -1, 0, 1, ..., 13, 14.

Let's count how many integers are there: $14 - (-19) - 1 = 14 + 19 - 1 = 33 - 1 = 32$. There are 34 integers from -19 to 14 (inclusive), and $14 - (-19) + 1 = 34$ integers.

Since $14 - (-20) - 1 = 14 + 20 - 1 = 33$. There are many integers between -20 and 15. For example, -19, -18, ..., 14. The number of integers is $14 - (-19) + 1 = 34$.

We need to find ten rational numbers. Since there are many integers between -20 and 15 (e.g., -19, -18, ..., 14), we can easily pick any ten of these integers as numerators with the denominator 24.

Ten rational numbers between $\frac{-20}{24}$ and $\frac{15}{24}$ are, for example:

$\frac{-19}{24}, \frac{-18}{24}, \frac{-17}{24}, \frac{-16}{24}, \frac{-15}{24}, \frac{-14}{24}, \frac{-13}{24}, \frac{-12}{24}, \frac{-11}{24}, \frac{-10}{24} $.

(You could pick any ten integers between -20 and 15).

If required, simplify these rational numbers to their standard form:

  • $\frac{-19}{24}$ (Standard form)
  • $\frac{-18}{24} = \frac{-3}{4}$ (Standard form)
  • $\frac{-17}{24}$ (Standard form)
  • $\frac{-16}{24} = \frac{-2}{3}$ (Standard form)
  • $\frac{-15}{24} = \frac{-5}{8}$ (Standard form)
  • $\frac{-14}{24} = \frac{-7}{12}$ (Standard form)
  • $\frac{-13}{24}$ (Standard form)
  • $\frac{-12}{24} = \frac{-1}{2}$ (Standard form)
  • $\frac{-11}{24}$ (Standard form)
  • $\frac{-10}{24} = \frac{-5}{12}$ (Standard form)

Thus, ten rational numbers between $\frac{-5}{6}$ and $\frac{5}{8}$ could be $\frac{-19}{24}, \frac{-3}{4}, \frac{-17}{24}, \frac{-2}{3}, \frac{-5}{8}, \frac{-7}{12}, \frac{-13}{24}, \frac{-1}{2}, \frac{-11}{24}, \frac{-5}{12}$.



Basic Algebraic Operations on Rational Numbers

Rational numbers are a broad category that includes all integers and fractions (both positive and negative). Just like other sets of numbers we have studied, we can perform the basic arithmetic operations – addition, subtraction, multiplication, and division – on rational numbers. The rules for these operations are extensions of the rules we learned for fractions and integers, with special attention needed for handling positive and negative signs.


1. Addition of Rational Numbers

To add two rational numbers, the process depends on whether they have the same denominator or different denominators. It's generally easiest if they have the same positive denominator.

(a) Addition of Rational Numbers with the Same Denominator:

If two rational numbers have the same positive denominator, their sum is found by adding their numerators and keeping the common denominator the same. The rules for adding integers apply to the numerators.

If $a, b$ are integers and $c$ is a positive integer, $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c} $.

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

Denominators are the same (7). Add the numerators (2 and 3): $2+3=5$. Keep the denominator 7.

$\frac{2}{7} + \frac{3}{7} = \frac{2+3}{7} = \frac{5}{7} $.

Example: Add $\frac{-5}{9}$ and $\frac{2}{9}$.

Denominators are the same (9). Add the numerators (-5 and 2): $-5+2$. Using integer addition rules, $|-5|=5, |2|=2$, difference $5-2=3$, sign of -5 is negative. $-5+2=-3$. Keep the denominator 9.

$\frac{-5}{9} + \frac{2}{9} = \frac{-5+2}{9} = \frac{-3}{9} $.

Simplify the resulting rational number $\frac{-3}{9}$ (HCF of 3 and 9 is 3): $\frac{-3 \div 3}{9 \div 3} = \frac{-1}{3}$.

So, $\frac{-5}{9} + \frac{2}{9} = \frac{-1}{3} $.

(b) Addition of Rational Numbers with Different Denominators:

If the denominators are different, or if one or both denominators are negative, the process is similar to adding unlike fractions:

  1. If any denominator is negative, convert the rational number to an equivalent one with a positive denominator (e.g., $\frac{a}{-b} = \frac{-a}{b}$).
  2. Find the Least Common Multiple (LCM) of the positive denominators. This will be the common denominator for adding.
  3. Convert each rational number into an equivalent rational number having the LCM as its denominator.
  4. Add the numerators of these new equivalent rational numbers using integer addition rules, and keep the common denominator.
  5. Simplify the resulting rational number to its standard form if possible.

Example: Add $\frac{-2}{3}$ and $\frac{3}{5}$.

Denominators are 3 and 5 (both positive). Find LCM(3, 5) = 15.

Convert $\frac{-2}{3}$ to an equivalent rational number with denominator 15: $\frac{-2 \times 5}{3 \times 5} = \frac{-10}{15}$.

Convert $\frac{3}{5}$ to an equivalent rational number with denominator 15: $\frac{3 \times 3}{5 \times 3} = \frac{9}{15}$.

Now add the equivalent rational numbers:

$\frac{-10}{15} + \frac{9}{15} = \frac{-10+9}{15} $.

Add the numerators: $-10+9$. Using integer addition, $|-10|=10, |9|=9$, difference $10-9=1$, sign of -10 is negative. $-10+9=-1$. Keep the denominator 15.

$ = \frac{-1}{15} $.

The resulting rational number $\frac{-1}{15}$ is in standard form.

Additive Inverse:

The Additive Inverse of a rational number $\frac{p}{q}$ is $-\frac{p}{q}$. When a rational number is added to its additive inverse, the sum is 0 (the additive identity). The additive inverse of $\frac{p}{q}$ can be written as $\frac{-p}{q}$ or $\frac{p}{-q}$.

Example: Additive inverse of $\frac{3}{4}$ is $-\frac{3}{4}$ (or $\frac{-3}{4}$). $\frac{3}{4} + (-\frac{3}{4}) = 0$.

Example: Additive inverse of $\frac{-5}{7}$ is $-(\frac{-5}{7}) = \frac{5}{7}$. $\frac{-5}{7} + \frac{5}{7} = 0$.


2. Subtraction of Rational Numbers

Subtraction of rational numbers is defined as adding the additive inverse of the rational number being subtracted. This converts any subtraction problem into an addition problem.

Rule: To subtract a rational number $\frac{c}{d}$ from another rational number $\frac{a}{b}$, add the additive inverse of $\frac{c}{d}$ to $\frac{a}{b}$.

$\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 $\frac{-c}{d}$), the rule becomes:

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

After converting subtraction to addition, follow the rules for adding rational numbers with different denominators (find LCM, convert to like rational numbers, add numerators).

Example: Subtract $\frac{1}{4}$ from $\frac{2}{3}$. This means calculate $\frac{2}{3} - \frac{1}{4}$.

Convert subtraction to addition: $\frac{2}{3} + (\text{additive inverse of } \frac{1}{4}) = \frac{2}{3} + \frac{-1}{4}$.

Denominators are 3 and 4. LCM(3, 4) = 12.

Convert to equivalent rational numbers with denominator 12:

Now add the equivalent rational numbers:

$\frac{8}{12} + \frac{-3}{12} = \frac{8 + (-3)}{12} = \frac{8-3}{12} = \frac{5}{12} $.

The result $\frac{5}{12}$ is in standard form.

Example: Subtract $\frac{-5}{7}$ from $\frac{-3}{8}$. This means calculate $\frac{-3}{8} - (\frac{-5}{7})$.

Convert subtraction to addition: $\frac{-3}{8} + (\text{additive inverse of } \frac{-5}{7})$. The additive inverse of $\frac{-5}{7}$ is $\frac{5}{7}$.

$\frac{-3}{8} - (\frac{-5}{7}) = \frac{-3}{8} + \frac{5}{7} $.

Denominators are 8 and 7. Find LCM(8, 7) = 56.

Convert to equivalent rational numbers with denominator 56:

Now add the equivalent rational numbers:

$\frac{-21}{56} + \frac{40}{56} = \frac{-21 + 40}{56} $.

Add the numerators using integer addition: $-21+40 = 40-21=19$. Keep the denominator 56.

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

The result $\frac{19}{56}$ is in standard form.


3. Multiplication of Rational Numbers

Multiplying rational numbers is similar to multiplying fractions. Multiply the numerators together to get the new numerator, and multiply the denominators together to get the new denominator. Pay careful attention to the signs of the numerators and denominators using the integer multiplication rules.

Rule: To multiply two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ (where $b \neq 0, d \neq 0$), multiply their numerators and multiply their denominators.

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

Remember the rules for signs in multiplication:

It is highly recommended to simplify by cancelling common factors between any numerator and any denominator before performing the multiplication, as it makes the calculation easier.

Example: Multiply $\frac{-3}{5} \times \frac{7}{8}$.

Multiply numerators $(-3) \times 7 = -21$. Multiply denominators $5 \times 8 = 40$.

$\frac{-3}{5} \times \frac{7}{8} = \frac{(-3) \times 7}{5 \times 8} = \frac{-21}{40} $.

The result $\frac{-21}{40}$ is in standard form.

Example: Multiply $\frac{9}{-5} \times \frac{-10}{27}$.

It's good practice to write rational numbers with positive denominators first, although not strictly necessary for multiplication.

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

The problem becomes $\frac{-9}{5} \times \frac{-10}{27}$. Multiply numerators $(-9) \times (-10) = 90$. Multiply denominators $5 \times 27 = 135$.

$\frac{-9}{5} \times \frac{-10}{27} = \frac{90}{135} $.

Simplify the resulting rational number $\frac{90}{135}$. HCF(90, 135) = 45. Divide numerator and denominator by 45.

$\frac{90 \div 45}{135 \div 45} = \frac{2}{3} $.

Using cancellation before multiplication is faster: $\frac{\cancel{-9}^{-1}}{\cancel{5}_1} \times \frac{\cancel{-10}^{-2}}{\cancel{27}_3} = \frac{(-1) \times (-2)}{1 \times 3} = \frac{2}{3}$.

So, $\frac{9}{-5} \times \frac{-10}{27} = \frac{2}{3} $.

Multiplicative Inverse (Reciprocal):

The Multiplicative Inverse or Reciprocal of a non-zero rational number $\frac{p}{q}$ is $\frac{q}{p}$. When a non-zero rational number is multiplied by its reciprocal, the product is 1 (the multiplicative identity).

Example: Reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$. $\frac{2}{3} \times \frac{3}{2} = 1$.

Example: Reciprocal of $\frac{-5}{11}$ is $\frac{11}{-5}$ or $\frac{-11}{5}$. $\frac{-5}{11} \times \frac{11}{-5} = \frac{-55}{-55} = 1$.

Zero (0 or $\frac{0}{1}$) does not have a reciprocal because $\frac{1}{0}$ is undefined.


4. Division of Rational Numbers

Dividing by a rational number is the same as multiplying by its multiplicative inverse (reciprocal). This is the key rule for division of rational numbers.

Rule: To divide one rational number $\frac{a}{b}$ by another non-zero rational number $\frac{c}{d}$ (the divisor), multiply the first rational number ($\frac{a}{b}$) by the reciprocal of the second rational number ($\frac{d}{c}$).

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

(where $\frac{c}{d} \neq 0$, which means $c \neq 0$ and $d \neq 0$)

After converting the division to multiplication, follow the rules for multiplication of rational numbers.

Example: Divide $\frac{-4}{5} \div \frac{2}{3}$.

The first rational number is $\frac{-4}{5}$. The divisor is $\frac{2}{3}$. Find the reciprocal of the divisor $\frac{2}{3}$, which is $\frac{3}{2}$.

Multiply the first rational number by the reciprocal of the second:

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

Multiply the numerators and denominators, using integer multiplication rules for signs:

$= \frac{(-4) \times 3}{5 \times 2} = \frac{-12}{10} $.

Simplify the resulting rational number $\frac{-12}{10}$ (HCF of 12 and 10 is 2): $\frac{-12 \div 2}{10 \div 2} = \frac{-6}{5}$.

Using cancellation before multiplication: $\frac{\cancel{-4}^{-2}}{5} \times \frac{3}{\cancel{2}_1} = \frac{-2 \times 3}{5 \times 1} = \frac{-6}{5}$.

So, $\frac{-4}{5} \div \frac{2}{3} = \frac{-6}{5} $.

Example: Divide $\frac{7}{-8} \div \frac{-3}{4}$.

First, write the rational number $\frac{7}{-8}$ with a positive denominator: $\frac{7}{-8} = \frac{7 \times (-1)}{-8 \times (-1)} = \frac{-7}{8}$.

The divisor is $\frac{-3}{4}$. Find its reciprocal: $\frac{4}{-3}$ or $\frac{-4}{3}$. Let's use $\frac{-4}{3}$.

The problem becomes $\frac{-7}{8} \div \frac{-3}{4}$. Multiply the first rational number by the reciprocal of the second:

$\frac{-7}{8} \div \frac{-3}{4} = \frac{-7}{8} \times \frac{-4}{3} $.

Multiply numerators and denominators, using integer rules for signs:

$= \frac{(-7) \times (-4)}{8 \times 3} = \frac{28}{24} $.

Simplify $\frac{28}{24}$ (HCF of 28 and 24 is 4): $\frac{28 \div 4}{24 \div 4} = \frac{7}{6}$.

Using cancellation: $\frac{-7}{\cancel{8}_2} \times \frac{\cancel{4}^1}{-3} = \frac{-7 \times 1}{2 \times (-3)} = \frac{-7}{-6} = \frac{7}{6}$.

So, let's emphasize the rule: multiply by the reciprocal $\frac{d}{c}$ where $c, d$ are the original numerator/denominator of the divisor.

So, $\frac{7}{-8} \div \frac{-3}{4} = \frac{7}{-8} \times \frac{4}{-3} = \frac{7 \times 4}{(-8) \times (-3)} = \frac{28}{24} = \frac{7}{6} $.


Properties of Operations on Rational Numbers (Brief Mention)

Rational numbers possess many properties under the basic operations, similar to integers and fractions.

Example 1. Find the sum: $\frac{-9}{10} + \frac{22}{15}$.

Answer:

We need to find the sum of $\frac{-9}{10}$ and $\frac{22}{15}$. These are rational numbers with different denominators.

Step 1: Ensure positive denominators (both are already 10 and 15, which are positive).

Step 2: Find the LCM of the denominators 10 and 15.

Multiples of 10: 10, 20, 30, ...

Multiples of 15: 15, 30, ...

LCM(10, 15) = 30.

Step 3: Convert both rational numbers to equivalent rational numbers with denominator 30.

  • For $\frac{-9}{10}$: Multiply numerator and denominator by 3 ($30 \div 10 = 3$).

    $\frac{-9}{10} = \frac{-9 \times 3}{10 \times 3} = \frac{-27}{30} $.

  • For $\frac{22}{15}$: Multiply numerator and denominator by 2 ($30 \div 15 = 2$).

    $\frac{22}{15} = \frac{22 \times 2}{15 \times 2} = \frac{44}{30} $.

Step 4: Add the numerators of these equivalent rational numbers and keep the common denominator.

Sum $= \frac{-27}{30} + \frac{44}{30} = \frac{-27 + 44}{30} $.

Add the numerators using integer addition rules: $-27 + 44$. Find the difference of absolute values: $44 - 27 = 17$. Sign of 44 (positive) is positive. So, $-27 + 44 = 17$.

Sum $= \frac{17}{30} $.

Step 5: Simplify. HCF(17, 30) = 1. The result $\frac{17}{30}$ is in standard form.

So, $\frac{-9}{10} + \frac{22}{15} = \frac{17}{30}$.

Example 2. Find: $\frac{7}{24} - \frac{17}{36}$.

Answer:

We need to find the difference $\frac{7}{24} - \frac{17}{36}$. Convert subtraction to addition of the additive inverse: $\frac{7}{24} + (-\frac{17}{36})$.

Denominators are 24 and 36 (both positive). Find LCM(24, 36).

$ \begin{array}{c|cc} 2 & 24 \; , & 36 \\ \hline 2 & 12 \; , & 18 \\ \hline 2 & 6 \; , & 9 \\ \hline 3 & 3 \; , & 9 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array} $

LCM = $2 \times 2 \times 2 \times 3 \times 3 = 8 \times 9 = 72$.

Convert to equivalent fractions with denominator 72:

  • For $\frac{7}{24}$: $72 \div 24 = 3$. Multiply by 3: $\frac{7 \times 3}{24 \times 3} = \frac{21}{72}$.
  • For $\frac{17}{36}$: $72 \div 36 = 2$. Multiply by 2: $\frac{17 \times 2}{36 \times 2} = \frac{34}{72}$. So, $-\frac{17}{36} = \frac{-34}{72}$.

Now perform the addition of the equivalent rational numbers:

Difference $= \frac{21}{72} + \frac{-34}{72} = \frac{21 + (-34)}{72} $.

Add the numerators using integer addition: $21 + (-34) = 21 - 34$. Find the difference of absolute values: $34-21=13$. Sign of -34 (negative) is negative. $21-34=-13$.

Difference $= \frac{-13}{72} $.

Simplify. HCF(13, 72) = 1. The result $\frac{-13}{72}$ is in standard form.

So, $\frac{7}{24} - \frac{17}{36} = \frac{-13}{72}$.

Example 3. Find the product: $\frac{-6}{5} \times \frac{9}{11}$.

Answer:

We need to find the product of $\frac{-6}{5}$ and $\frac{9}{11}$.

Multiply the numerators: $(-6) \times 9$. Negative $\times$ Positive $=$ Negative. $6 \times 9 = 54$. So, $(-6) \times 9 = -54$.

Multiply the denominators: $5 \times 11 = 55$.

Product $= \frac{(-6) \times 9}{5 \times 11} = \frac{-54}{55} $.

Simplify. HCF(54, 55) = 1. The result $\frac{-54}{55}$ is in standard form.

So, $\frac{-6}{5} \times \frac{9}{11} = \frac{-54}{55}$.

Example 4. Find the value of: $\frac{-4}{1} \div \frac{2}{3}$. (Note: -4 is the integer -4, which can be written as $\frac{-4}{1}$).

Answer:

We need to perform the division $\frac{-4}{1} \div \frac{2}{3}$.

To divide by a rational number ($\frac{2}{3}$), we multiply the first rational number ($\frac{-4}{1}$) by the reciprocal of the second rational number ($\frac{2}{3}$).

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Division $= \frac{-4}{1} \times \text{Reciprocal of } \frac{2}{3} = \frac{-4}{1} \times \frac{3}{2} $.

Multiply the numerators and denominators:

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

Simplify the resulting rational number $\frac{-12}{2}$ (HCF of 12 and 2 is 2): $\frac{-12 \div 2}{2 \div 2} = \frac{-6}{1} = -6$.

Using cancellation before multiplication: $\frac{\cancel{-4}^{-2}}{1} \times \frac{3}{\cancel{2}_1} = \frac{-2 \times 3}{1 \times 1} = \frac{-6}{1} = -6$.

So, $\frac{-4}{1} \div \frac{2}{3} = -6 $.