Class 8
Mathematics
Federal Board of Intermediate and Secondary Education (FBISE) (FBISE)
📚 Model Textbook of Mathematics Grade 8 (FBISE / NBF)
Class 8 Mathematics - Ch 1: Real Numbers, Decimal Classification, Properties, Inequalities & Absolute Values (FBISE)
📖 Chapter 1: Real Numbers
📅 Updated: Sep 13, 2026
Teacher & Student Roadmap
Grade 8 Mathematics • FBISE / National Curriculum (NBF)
Instructional Guide: Unit 01 Real Numbers (حقیقی اعداد)
Target Student Learning Outcomes (SLOs)
- Define and differentiate Rational numbers ($\mathbb{Q}$) and Irrational numbers ($\mathbb{Q}'$).
- Understand the set of Real Numbers ($\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}'$) and disjoint property ($\mathbb{Q} \cap \mathbb{Q}' = \emptyset$).
- Convert rational fractions into terminating and recurring (periodic) decimal fractions using dot ($\dot{a}$) and bar ($\bar{a}$) notation.
- Recognize non-terminating and non-recurring decimals as irrational numbers (e.g. $\pi, \sqrt{2}, \sqrt{3}, \sqrt{5}$).
- Represent rational and irrational numbers on the number line geometrically using the Pythagorean theorem ($OA^2 + AB^2 = OB^2$) and a compass.
- Verify and apply Real Number properties: Closure, Commutative, Associative, Identity, Inverse, and Distributive properties w.r.t addition and multiplication.
- Understand Ordering properties of Real Numbers (Reflexive, Symmetric, Transitive, Additive, Multiplicative, Cancellation, Trichotomy, and Inequality Inverses).
- Define and evaluate Absolute Value ($|x|$) and measure distances between points on a number line ($|a - b|$).
- Solve multi-step real-world word problems involving rational numbers and decimals.
Prerequisites & Bridge Concepts
- Class 7 Integers & Fractions: Arithmetic operations on positive and negative integers, LCM, equivalent fractions.
- Class 7 Rational Numbers: Form $\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$.
- Pythagoras Theorem: $c = \sqrt{a^2 + b^2}$ for right-angled triangles.
- Number Line Basics: Positive numbers to the right of zero, negative numbers to the left of zero.
Common Misconceptions & Pitfalls
- Value of $\pi$ vs $\frac{22}{7}$: $\pi$ is an irrational number (its true decimal $3.14159265...$ never terminates and never repeats). The fraction $\frac{22}{7} = 3.\overline{142857}$ is a rational approximation used for practical calculations.
- Multiplying Inequalities by a Negative Number: Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign! For example, $3 < 5 \implies 3 \times (-2) > 5 \times (-2) \implies -6 > -10$.
- Multiplicative Inverse of Zero: The multiplicative inverse of $0$ does NOT exist because division by zero ($\frac{1}{0}$) is undefined.
- Absolute Value Output: Absolute value $|x|$ represents geometric distance and is always non-negative ($|x| \ge 0$). $|-7| = 7$, never $-7$.
Pedagogical Flow & Practical Teaching Tips
Start with a physical number line. Review Natural numbers ($\mathbb{N}$), Whole numbers ($\mathbb{W}$), and Integers ($\mathbb{Z}$). Show that gaps between integers contain Rational numbers ($\mathbb{Q}$). Then introduce the question: "Can every point on the line be written as a fraction?" Introduce $\sqrt{2}$ via a unit square diagonal to prove the existence of gaps that only Irrational numbers ($\mathbb{Q}'$) can fill. Combine both sets to establish the continuum of Real Numbers ($\mathbb{R}$).
Kid-Friendly Rhymes & Golden Math Rules
Rational vs. Irrational Rhyme:
"If it stops or loops in a repeating track,
It's a Rational Number, clean and exact!
If it wanders on forever with no repeat in sight,
It's an Irrational Number, mysterious and bright!"
"If it stops or loops in a repeating track,
It's a Rational Number, clean and exact!
If it wanders on forever with no repeat in sight,
It's an Irrational Number, mysterious and bright!"
Inequality Flip Warning:
"Multiply or divide by a minus sign,
Flip the crocodile mouth every single time!
Less becomes greater, greater becomes less,
Keep this golden rule to guarantee success!"
"Multiply or divide by a minus sign,
Flip the crocodile mouth every single time!
Less becomes greater, greater becomes less,
Keep this golden rule to guarantee success!"
Absolute Value Motto:
"Distance from zero, whether left or right,
Absolute value makes all numbers bright!
Strip away the negative, keep the size true,
Non-negative distance is all it gives to you!"
"Distance from zero, whether left or right,
Absolute value makes all numbers bright!
Strip away the negative, keep the size true,
Non-negative distance is all it gives to you!"
Key Definitions & Properties Reference Bank
| Property / Concept | Addition Form ($+$) | Multiplication Form ($\times$) | Example / Meaning |
|---|---|---|---|
| Closure Property | $a + b \in \mathbb{R}$ | $a \cdot b \in \mathbb{R}$ | $5 + 3 = 8 \in \mathbb{R}$; $5 \times 3 = 15 \in \mathbb{R}$ |
| Commutative Property | $a + b = b + a$ | $a \cdot b = b \cdot a$ | $5 + 8 = 8 + 5 = 13$; $5 \times 8 = 8 \times 5 = 40$ |
| Associative Property | $a + (b + c) = (a + b) + c$ | $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ | $3 + (5 + 7) = (3 + 5) + 7 = 15$; $3 \times (5 \times 7) = 105$ |
| Identity Property | $a + 0 = 0 + a = a$ (Additive: $0$) | $a \cdot 1 = 1 \cdot a = a$ (Multiplicative: $1$) | $5 + 0 = 5$; $5 \times 1 = 5$ |
| Inverse Property | $a + (-a) = -a + a = 0$ | $a \cdot \frac{1}{a} = \frac{1}{a} \cdot a = 1$ ($a \neq 0$) | $12 + (-12) = 0$; $12 \times \frac{1}{12} = 1$ |
| Distributive Property | $a(b + c) = ab + ac \quad \text{and} \quad a(b - c) = ab - ac$ | $4(5 + 8) = 4(5) + 4(8) = 20 + 32 = 52$ | |
| Absolute Value | $|x| = x \text{ if } x \ge 0, \quad |x| = -x \text{ if } x < 0$ | $|-5| = 5$, $|+5| = 5$, $|0| = 0$. Distance between $a, b$ is $|a - b|$. | |
Classification of Real Numbers ($\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}'$)
Figure 1.1: Hierarchy of the Real Number System. Notice that $\mathbb{Q}$ and $\mathbb{Q}'$ have no common elements ($\mathbb{Q} \cap \mathbb{Q}' = \emptyset$).
Geometric Construction of $\sqrt{2}$ on the Number Line
Figure 1.2: Construction of $\sqrt{2}$ using Pythagoras Theorem: $OB = \sqrt{OA^2 + AB^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$. The compass arc from $B$ intersects the line at point $P = \sqrt{2}$.
1.1 Real Numbers Hierarchy & Subsets
In mathematics, numbers evolved step by step to solve increasingly complex problems:
- Natural Numbers ($\mathbb{N}$): Counting numbers used in daily life: $\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}$.
- Whole Numbers ($\mathbb{W}$): Natural numbers including zero: $\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}$. ($\mathbb{N} \subset \mathbb{W}$).
- Integers ($\mathbb{Z}$): Whole numbers along with negative counting numbers: $\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}$. ($\mathbb{W} \subset \mathbb{Z}$).
- Rational Numbers ($\mathbb{Q}$): Numbers that can be written in the fractional form $\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$: $$\mathbb{Q} = \left\{x : x = \frac{a}{b}, \text{ where } a, b \in \mathbb{Z} \text{ and } b \neq 0\right\}$$ Every integer is rational because $n = \frac{n}{1}$. Thus, $\mathbb{Z} \subset \mathbb{Q}$.
- Irrational Numbers ($\mathbb{Q}'$): Numbers that cannot be written in the form $\frac{a}{b}$ for integers $a, b$ ($b \neq 0$). When expressed as decimals, they are non-terminating and non-recurring: $$\mathbb{Q}' = \{x : x \text{ cannot be expressed as } \frac{a}{b}, a, b \in \mathbb{Z}, b \neq 0\}$$ Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \pi \approx 3.14159\dots$
- Real Numbers ($\mathbb{R}$): The complete set formed by uniting all rational and irrational numbers: $$\mathbb{R} = \mathbb{Q} \cup \mathbb{Q}' \quad \text{and} \quad \mathbb{Q} \cap \mathbb{Q}' = \emptyset$$ The rational and irrational sets are disjoint (they share no elements) and exhaustive (together they form all real numbers).
1.2 Decimal Representation of Real Numbers
Every real number can be written in decimal form. There are three fundamental categories:
1. Terminating Decimals ($\mathbb{Q}$)
Decimals with a finite number of digits after the decimal point (division comes to an end with remainder $0$).
• 3/5 = 0.6
• 13/20 = 0.65
• 9/8 = 1.125
• 1.2578
• 13/20 = 0.65
• 9/8 = 1.125
• 1.2578
2. Non-Terminating Recurring Decimals ($\mathbb{Q}$)
Decimals that never end, but a single digit or a block of digits repeats periodically. Shown by dots or bars.
• 1/3 = 0.333... = 0.3̇ = 0.3̄
• 3 2/11 = 3.1818... = 3.1̇8̇ = 3.18̄
• 4/7 = 0.571428571428... = 0.571428̄
• 7/11 = 0.636363... = 0.63̄
• 3 2/11 = 3.1818... = 3.1̇8̇ = 3.18̄
• 4/7 = 0.571428571428... = 0.571428̄
• 7/11 = 0.636363... = 0.63̄
3. Non-Terminating Non-Recurring Decimals ($\mathbb{Q}'$)
Decimals that go on infinitely without repeating in any periodic sequence. These represent Irrational Numbers.
• √2 = 1.4142135623...
• √3 = 1.7320508075...
• π = 3.1415926535...
• √17 = 4.123105626...
• √3 = 1.7320508075...
• π = 3.1415926535...
• √17 = 4.123105626...
1.3 In-Text Check Points Solved in Full
Check Point (Book Page 8): Represent $\sqrt{7}$ and $\sqrt{11}$ on Number Line
Part 1: Construction of $\sqrt{7}$ on Number Line
- First construct $\sqrt{4} = 2$. On the number line, mark $OA = 2$ units.
- Draw a perpendicular segment $AB = 1$ unit at $A$. Join $OB$. By Pythagoras Theorem: $$OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$
- At point $B$, draw a perpendicular line segment $BC = 1$ unit to $OB$. Join $OC$: $$OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{5 + 1} = \sqrt{6}$$
- At point $C$, draw a perpendicular segment $CD = 1$ unit to $OC$. Join $OD$: $$OD = \sqrt{OC^2 + CD^2} = \sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{6 + 1} = \sqrt{7}$$
- With center at $O(0)$ and radius equal to $OD = \sqrt{7} \approx 2.646$, draw an arc with a compass to cut the positive number line at point $P$. Point $P$ represents $\sqrt{7}$.
Part 2: Construction of $\sqrt{11}$ on Number Line
- Start at $OA = 3$ units ($\sqrt{9} = 3$).
- Draw perpendicular $AB = 1$ unit at $A$. Then $OB = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.
- Draw perpendicular $BC = 1$ unit to $OB$. Then $OC = \sqrt{(\sqrt{10})^2 + 1^2} = \sqrt{10 + 1} = \sqrt{11}$.
- With center $O$ and radius $OC = \sqrt{11} \approx 3.317$, draw an arc intersecting the number line at point $Q$. Point $Q$ represents $\sqrt{11}$.
1.4 Exercise 1.1 — 100% Step-by-Step Solved
Exercise 1.1 Q1: Separate Rational and Irrational Numbers
Identify whether each given number is Rational ($\mathbb{Q}$) or Irrational ($\mathbb{Q}'$) with mathematical reasoning:
| Part | Number | Classification | Detailed Reason |
|---|---|---|---|
| (i) | $\sqrt{9}$ | Rational | $\sqrt{9} = 3 = \frac{3}{1}$, which is a terminating integer in $\frac{a}{b}$ form. |
| (ii) | $12$ | Rational | $12 = \frac{12}{1}$, an integer expressed as a fraction ($a,b \in \mathbb{Z}, b \neq 0$). |
| (iii) | $\frac{5}{9}$ | Rational | Written in standard fractional form $\frac{a}{b} = 0.555\dots = 0.\bar{5}$ (recurring decimal). |
| (iv) | $\sqrt{8}$ | Irrational | $\sqrt{8} = 2\sqrt{2} \approx 2.828427\dots$, non-terminating and non-recurring decimal. |
| (v) | $\sqrt{100}$ | Rational | $\sqrt{100} = 10 = \frac{10}{1}$, perfect square root resulting in an integer. |
| (vi) | $\frac{13}{2}$ | Rational | $\frac{13}{2} = 6.5$, a terminating decimal fraction. |
| (vii) | $\sqrt{126}$ | Irrational | $126$ is not a perfect square ($\sqrt{126} = 3\sqrt{14} \approx 11.22497\dots$), non-terminating non-recurring. |
| (viii) | $\frac{25}{9}$ | Rational | $\frac{25}{9} = 2.777\dots = 2.\bar{7}$, non-terminating recurring decimal. |
| (ix) | $\sqrt{169}$ | Rational | $\sqrt{169} = 13 = \frac{13}{1}$, an integer ($169 = 13^2$). |
| (x) | $\sqrt{26}$ | Irrational | $26$ is not a perfect square ($\sqrt{26} \approx 5.099019\dots$), non-terminating non-recurring. |
Exercise 1.1 Q2: Convert to Decimal & State Terminating or Non-Terminating
(i) $\frac{4}{9}$
Long Division: $4 \div 9 = 0.4444\dots = 0.\bar{4}$
Non-Terminating (Recurring)
Long Division: $4 \div 9 = 0.4444\dots = 0.\bar{4}$
Non-Terminating (Recurring)
(ii) $\frac{13}{20}$
Division: $\frac{13 \times 5}{20 \times 5} = \frac{65}{100} = 0.65$
Terminating
Division: $\frac{13 \times 5}{20 \times 5} = \frac{65}{100} = 0.65$
Terminating
(iii) $\frac{1}{6}$
Long Division: $1 \div 6 = 0.1666\dots = 0.1\bar{6}$
Non-Terminating (Recurring)
Long Division: $1 \div 6 = 0.1666\dots = 0.1\bar{6}$
Non-Terminating (Recurring)
(iv) $\frac{7}{3}$
Long Division: $7 \div 3 = 2.3333\dots = 2.\bar{3}$
Non-Terminating (Recurring)
Long Division: $7 \div 3 = 2.3333\dots = 2.\bar{3}$
Non-Terminating (Recurring)
(v) $\frac{9}{8}$
Division: $\frac{9 \times 125}{8 \times 125} = \frac{1125}{1000} = 1.125$
Terminating
Division: $\frac{9 \times 125}{8 \times 125} = \frac{1125}{1000} = 1.125$
Terminating
(vi) $\frac{13}{8}$
Division: $\frac{13 \times 125}{8 \times 125} = \frac{1625}{1000} = 1.625$
Terminating
Division: $\frac{13 \times 125}{8 \times 125} = \frac{1625}{1000} = 1.625$
Terminating
(vii) $\frac{11}{15}$
Long Division: $11 \div 15 = 0.7333\dots = 0.7\bar{3}$
Non-Terminating (Recurring)
Long Division: $11 \div 15 = 0.7333\dots = 0.7\bar{3}$
Non-Terminating (Recurring)
(viii) $\frac{7}{11}$
Long Division: $7 \div 11 = 0.636363\dots = 0.\overline{63}$
Non-Terminating (Recurring)
Long Division: $7 \div 11 = 0.636363\dots = 0.\overline{63}$
Non-Terminating (Recurring)
Exercise 1.1 Q3: Name the Type of Decimals (Rational or Irrational)
| Part | Decimal Number | Decimal Type | Rational / Irrational |
|---|---|---|---|
| (i) | $1.2578$ | Terminating Decimal | Rational Number ($\in \mathbb{Q}$) |
| (ii) | $0.33333\dots$ | Non-Terminating Recurring Decimal | Rational Number ($\in \mathbb{Q}$) |
| (iii) | $1.3414213662\dots$ | Non-Terminating Non-Recurring Decimal | Irrational Number ($\in \mathbb{Q}'$) |
| (iv) | $5.142857142857\dots$ | Non-Terminating Recurring Decimal (period $142857$) | Rational Number ($\in \mathbb{Q}$) |
| (v) | $2.236067977\dots$ | Non-Terminating Non-Recurring Decimal ($\sqrt{5}$) | Irrational Number ($\in \mathbb{Q}'$) |
| (vi) | $4.36363636\dots$ | Non-Terminating Recurring Decimal (period $36$) | Rational Number ($\in \mathbb{Q}$) |
| (vii) | $4.123105626\dots$ | Non-Terminating Non-Recurring Decimal ($\sqrt{17}$) | Irrational Number ($\in \mathbb{Q}'$) |
Exercise 1.1 Q4: Word Problem on Fractions (Savings Calculation)
Problem Statement:
Ismah uses $\frac{1}{4}$ of her monthly pay cheque on utility bills and $\frac{1}{10}$ of remaining for savings. If her last pay cheque was $\text{Rs. } 6000.80$, how much did she put in savings?
Step-by-step Solution:
1. Total monthly pay cheque $= \text{Rs. } 6000.80$
2. Amount spent on utility bills $= \frac{1}{4} \times 6000.80 = \text{Rs. } 1500.20$
3. Remaining amount $= 6000.80 - 1500.20 = \text{Rs. } 4500.60$
(Alternatively: Fraction remaining $= 1 - \frac{1}{4} = \frac{3}{4}$; $\frac{3}{4} \times 6000.80 = \text{Rs. } 4500.60$)
4. Amount put into savings $= \frac{1}{10} \text{ of remaining} = \frac{1}{10} \times 4500.60 = \text{Rs. } 450.06$
Final Answer:
Ismah put exactly $\text{Rs. } 450.06$ in savings.
1. Total monthly pay cheque $= \text{Rs. } 6000.80$
2. Amount spent on utility bills $= \frac{1}{4} \times 6000.80 = \text{Rs. } 1500.20$
3. Remaining amount $= 6000.80 - 1500.20 = \text{Rs. } 4500.60$
(Alternatively: Fraction remaining $= 1 - \frac{1}{4} = \frac{3}{4}$; $\frac{3}{4} \times 6000.80 = \text{Rs. } 4500.60$)
4. Amount put into savings $= \frac{1}{10} \text{ of remaining} = \frac{1}{10} \times 4500.60 = \text{Rs. } 450.06$
Final Answer:
Ismah put exactly $\text{Rs. } 450.06$ in savings.
Exercise 1.1 Q5: Word Problem on Ribbon Length Left Over
Problem Statement:
Raiqa is decorating for a birthday party and she purchase $20.7\text{ feet}$ of purple ribbon. She plans to use $\frac{1}{3}$ of the ribbon in the drawing room and $\frac{1}{2}$ of remaining in the dining room. How much ribbon will she have left over?
Step-by-step Solution:
1. Total length of ribbon $= 20.7\text{ feet}$
2. Ribbon used in drawing room $= \frac{1}{3} \times 20.7 = 6.9\text{ feet}$
3. Remaining ribbon after drawing room $= 20.7 - 6.9 = 13.8\text{ feet}$
4. Ribbon used in dining room $= \frac{1}{2} \text{ of remaining} = \frac{1}{2} \times 13.8 = 6.9\text{ feet}$
5. Ribbon left over $= 13.8 - 6.9 = 6.9\text{ feet}$
Final Answer:
Raiqa will have $6.9\text{ feet}$ of ribbon left over.
1. Total length of ribbon $= 20.7\text{ feet}$
2. Ribbon used in drawing room $= \frac{1}{3} \times 20.7 = 6.9\text{ feet}$
3. Remaining ribbon after drawing room $= 20.7 - 6.9 = 13.8\text{ feet}$
4. Ribbon used in dining room $= \frac{1}{2} \text{ of remaining} = \frac{1}{2} \times 13.8 = 6.9\text{ feet}$
5. Ribbon left over $= 13.8 - 6.9 = 6.9\text{ feet}$
Final Answer:
Raiqa will have $6.9\text{ feet}$ of ribbon left over.
1.5 Exercise 1.2 — Properties of Real Numbers & Absolute Value (100% Solved)
Exercise 1.2 Q1: Name the Properties Used in the Following Equations
| Part | Equation | Property Used | Standard Algebraic Formula |
|---|---|---|---|
| (i) | $7 + 3 = 3 + 7$ | Commutative property w.r.t addition | $a + b = b + a$ |
| (ii) | $(p + q) + r = p + (q + r)$ | Associative property w.r.t addition | $(a + b) + c = a + (b + c)$ |
| (iii) | $(\sqrt{3} \times \sqrt{5}) \times \sqrt{7} = \sqrt{3} \times (\sqrt{5} \times \sqrt{7})$ | Associative property w.r.t multiplication | $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ |
| (iv) | $80 + 0 = 0 + 80 = 80$ | Additive Identity property | $a + 0 = 0 + a = a$ |
| (v) | $100 \times 1 = 100$ | Multiplicative Identity property | $a \cdot 1 = 1 \cdot a = a$ |
| (vi) | $p + (-p) = 0$ | Additive Inverse property | $a + (-a) = 0$ |
| (vii) | $\sqrt{3} \times \sqrt{3} = 3$ | Closure property w.r.t multiplication / Definition of square root | $\sqrt{a} \cdot \sqrt{a} = a \in \mathbb{R}$ |
| (viii) | $\frac{1}{\sqrt{7}} \times \sqrt{7} = 1$ | Multiplicative Inverse property | $a \cdot \frac{1}{a} = 1$ ($a \neq 0$) |
| (ix) | $\sqrt{3} \times \sqrt{7} = \sqrt{7} \times \sqrt{3}$ | Commutative property w.r.t multiplication | $a \cdot b = b \cdot a$ |
| (x) | $2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$ | Distributive property of multiplication over addition | $(2 + 3)\sqrt{2} = 5\sqrt{2}$ |
Exercise 1.2 Q2: Find the Additive Inverse of the Following
Rule: The additive inverse of $x$ is $-x$ such that $x + (-x) = 0$.
(i) $\frac{12}{5}$ → $-\frac{12}{5}$
(ii) $\frac{-6}{13}$ → $\frac{6}{13}$
(iii) $\frac{27}{5}$ → $-\frac{27}{5}$
(iv) $\frac{8}{-5}$ → $\frac{8}{5}$
(v) $13\frac{7}{8} = \frac{111}{8}$ → $-13\frac{7}{8} = -\frac{111}{8}$
(vi) $\frac{-5}{17}$ → $\frac{5}{17}$
(vii) $-4\frac{3}{6} = -\frac{9}{2}$ → $4\frac{3}{6} = \frac{9}{2}$
(viii) $\frac{3}{11}$ → $-\frac{3}{11}$
Exercise 1.2 Q3: Find the Multiplicative Inverse (Reciprocal)
Rule: The multiplicative inverse of $x \neq 0$ is $\frac{1}{x}$ such that $x \cdot \frac{1}{x} = 1$.
(i) $-\frac{4}{3}$ → $-\frac{3}{4}$
(ii) $\frac{1}{2}$ → $2$
(iii) $\frac{1}{\sqrt{7}}$ → $\sqrt{7}$
(iv) $\frac{-5}{-7} = \frac{5}{7}$ → $\frac{7}{5}$
(v) $8 + 3 = 11$ → $\frac{1}{11}$
(vi) $-1$ → $-1$ (self inverse)
(vii) $9 - 7 = 2$ → $\frac{1}{2}$
(viii) $a$ ($a \neq 0$) → $\frac{1}{a}$
Exercise 1.2 Q4: Identify the Property that Justifies Each Statement
| Part | Statement | Justifying Property |
|---|---|---|
| (i) | $1 \times (y - 2) = 1 \times y - 1 \times 2$ | Distributive Property of Multiplication over Subtraction |
| (ii) | $(0.2) \cdot 5 = 1$ (since $0.2 = \frac{1}{5}$) | Multiplicative Inverse Property |
| (iii) | $(x + 2) + y = y + (x + 2)$ | Commutative Property w.r.t Addition |
| (iv) | $-(3b) + (3b) = 0$ | Additive Inverse Property |
| (v) | $(x + 5) - 1 = x + (5 - 1)$ | Associative Property w.r.t Addition ($x + 5 + (-1) = x + [5 + (-1)]$) |
| (vi) | $-3(2 - y) = -6 + 3y$ | Distributive Property of Multiplication over Subtraction |
Exercise 1.2 Q5: Properties of Equality & Inequality
| Part | Statement | Name of Property | Explanation |
|---|---|---|---|
| (i) | $7 > 5 \implies 11 > 9$ | Additive property of inequality | Adding $+4$ to both sides ($7 + 4 > 5 + 4$) preserves inequality. |
| (ii) | $-5 < 3 \implies -1 < 7$ | Additive property of inequality | Adding $+4$ to both sides ($-5 + 4 < 3 + 4$). |
| (iii) | $6 < 8 \implies 36 < 48$ | Multiplicative property of inequality | Multiplying both sides by positive number $6 > 0$ ($6 \times 6 < 8 \times 6$). |
| (iv) | $p < 0 \implies -p > 0$ | Multiplicative property of inequality | Multiplying both sides by $-1 < 0$ reverses the inequality symbol. |
| (v) | $\frac{3}{5}x = \frac{3}{5}y \implies x = y$ | Cancellation property w.r.t multiplication | Cancelling the common non-zero multiplier $\frac{3}{5}$. |
| (vi) | $7x = 7y \implies 7y = 7x$ | Symmetric property of equality | If $A = B$ then $B = A$. |
| (vii) | $x + 5 = y + 5 \text{ and } y + 5 = z + 5 \implies x + 5 = z + 5$ | Transitive property of equality | If $A = B$ and $B = C$ then $A = C$. |
| (viii) | $x + t = y + t \implies x = y$ | Cancellation property w.r.t addition | Cancelling identical added term $t$ from both sides. |
| (ix) | $p = q \implies -\frac{2}{3}p = -\frac{2}{3}q$ | Multiplicative property of equality | Multiplying both sides of an equality by $-\frac{2}{3}$. |
| (x) | $a > b \implies \frac{1}{b} > \frac{1}{a} \text{ (i.e. } \frac{1}{a} < \frac{1}{b}\text{)}$ | Inequality multiplicative inverse property | Taking reciprocals of positive numbers reverses inequality order ($5 > 2 \implies \frac{1}{5} < \frac{1}{2}$). |
Exercise 1.2 Q6: Solve Absolute Value Expressions
(i) $|-7| + |5| - |9|$
$= 7 + 5 - 9$
$= 12 - 9$
$= \mathbf{3}$
$= 7 + 5 - 9$
$= 12 - 9$
$= \mathbf{3}$
(ii) $|-3 - 4| \times |-\frac{1}{4}| + 1$
$= |-7| \times \frac{1}{4} + 1$
$= 7 \times \frac{1}{4} + 1 = \frac{7}{4} + \frac{4}{4} = \mathbf{\frac{11}{4} = 2.75}$
$= |-7| \times \frac{1}{4} + 1$
$= 7 \times \frac{1}{4} + 1 = \frac{7}{4} + \frac{4}{4} = \mathbf{\frac{11}{4} = 2.75}$
(iii) $|-7 + 2| - |5 - 9| + |-4|$
$= |-5| - |-4| + |-4|$
$= 5 - 4 + 4 = \mathbf{5}$
$= |-5| - |-4| + |-4|$
$= 5 - 4 + 4 = \mathbf{5}$
(iv) $(|-1 + 4| + |-5 + 1|) \div |15 - 1|$
$= (|3| + |-4|) \div |14|$
$= (3 + 4) \div 14 = 7 \div 14 = \mathbf{\frac{1}{2} = 0.5}$
$= (|3| + |-4|) \div |14|$
$= (3 + 4) \div 14 = 7 \div 14 = \mathbf{\frac{1}{2} = 0.5}$
1.6 Review Exercise 1 — Complete Master Solutions
Review Exercise 1 Q1: Multiple Choice Questions (MCQs)
(i) Rational numbers between 8 and 9:
Correct Option: (a) infinite (Dense property of rational numbers).
Correct Option: (a) infinite (Dense property of rational numbers).
(ii) Fraction $\frac{5}{21}$ in decimal form:
$5 \div 21 = 0.238095\dots \approx 0.24$
Correct Option: (d) 0.24 (or (b) 0.23 truncated).
$5 \div 21 = 0.238095\dots \approx 0.24$
Correct Option: (d) 0.24 (or (b) 0.23 truncated).
(iii) Fraction $\frac{5}{12}$ as recurring decimal:
$5 \div 12 = 0.41666\dots = 0.41\bar{6}$
Correct Option: (a) $0.41\dot{6}$
$5 \div 12 = 0.41666\dots = 0.41\bar{6}$
Correct Option: (a) $0.41\dot{6}$
(iv) Additive inverse of $\frac{-3}{5}$:
$-\left(-\frac{3}{5}\right) = \frac{3}{5}$
Correct Option: (d) $\frac{3}{5}$
$-\left(-\frac{3}{5}\right) = \frac{3}{5}$
Correct Option: (d) $\frac{3}{5}$
(v) Reciprocal of $3\frac{4}{5}$:
$3\frac{4}{5} = \frac{19}{5} \implies \text{Reciprocal} = \frac{5}{19}$
Correct Option: (d) $\frac{5}{19}$
$3\frac{4}{5} = \frac{19}{5} \implies \text{Reciprocal} = \frac{5}{19}$
Correct Option: (d) $\frac{5}{19}$
(vi) If $\frac{4}{7}$ and $b$ are multiplicative inverses:
$\frac{4}{7} \times b = 1 \implies b = \frac{7}{4}$
Correct Option: (a) $\frac{7}{4}$
$\frac{4}{7} \times b = 1 \implies b = \frac{7}{4}$
Correct Option: (a) $\frac{7}{4}$
(vii) If $a \times b = 1$ and $b = 1\frac{2}{3} = \frac{5}{3}$:
$a = \frac{1}{b} = \frac{3}{5}$
Correct Option: (d) $\frac{3}{5}$
$a = \frac{1}{b} = \frac{3}{5}$
Correct Option: (d) $\frac{3}{5}$
(viii) If $a + b = 0$ and $a = \frac{-8}{-9} = \frac{8}{9}$:
$b = -a = -\frac{8}{9}$
Correct Option: (a) $-\frac{8}{9}$
$b = -a = -\frac{8}{9}$
Correct Option: (a) $-\frac{8}{9}$
(ix) Multiplicative inverse does not exist for:
Correct Option: (a) 0 (division by 0 is undefined).
Correct Option: (a) 0 (division by 0 is undefined).
(x) Property in $\frac{3}{4} + \frac{1}{2} = \frac{1}{2} + \frac{3}{4}$:
Correct Option: (c) commutative
Correct Option: (c) commutative
(xi) Which number is self multiplicative inverse?
$\frac{1}{-1} = -1$
Correct Option: (d) $-1$ (also $+1$).
$\frac{1}{-1} = -1$
Correct Option: (d) $-1$ (also $+1$).
(xii) $\left(-\frac{7}{6}\right) + \left(-\frac{2}{3}\right)$ is equal to:
$-\frac{7}{6} - \frac{4}{6} = -\frac{11}{6}$
Correct Option: (d) $-\frac{11}{6}$
$-\frac{7}{6} - \frac{4}{6} = -\frac{11}{6}$
Correct Option: (d) $-\frac{11}{6}$
Review Exercise 1 Q2: Decimal Form & Classification
| Number | Decimal Value | Type | Rational / Irrational |
|---|---|---|---|
| (i) $\frac{35}{37}$ | $0.\overline{945}$ | Non-terminating recurring | Rational |
| (ii) $\frac{8}{11}$ | $0.\overline{72}$ | Non-terminating recurring | Rational |
| (iii) $-\frac{\pi}{7}$ | $-0.44879895\dots$ | Non-terminating non-recurring | Irrational |
| (iv) $\frac{1}{\sqrt{8}}$ | $0.35355339\dots$ | Non-terminating non-recurring | Irrational |
| (v) $\frac{1}{\sqrt{4}}$ | $\frac{1}{2} = 0.5$ | Terminating decimal | Rational |
Review Exercise 1 Q3: Verify Distributive Property of Multiplication over Subtraction
Verify $x \times (y - z) = x \times y - x \times z$ for $x = \frac{2}{3}, y = -\frac{3}{4}, z = \frac{2}{5}$.
Left Hand Side (LHS):
$$\text{LHS} = x \times (y - z) = \frac{2}{3} \times \left(-\frac{3}{4} - \frac{2}{5}\right)$$ Take LCM of denominators $4$ and $5$, which is $20$:
$$-\frac{3}{4} - \frac{2}{5} = \frac{-15 - 8}{20} = -\frac{23}{20}$$ Multiply by $\frac{2}{3}$:
$$\text{LHS} = \frac{2}{3} \times \left(-\frac{23}{20}\right) = \frac{1 \times (-23)}{3 \times 10} = -\frac{23}{30}$$ Right Hand Side (RHS): $$\text{RHS} = (x \times y) - (x \times z) = \left(\frac{2}{3} \times -\frac{3}{4}\right) - \left(\frac{2}{3} \times \frac{2}{5}\right)$$ $$\left(\frac{2}{3} \times -\frac{3}{4}\right) = -\frac{2}{4} = -\frac{1}{2}$$ $$\left(\frac{2}{3} \times \frac{2}{5}\right) = \frac{4}{15}$$ $$\text{RHS} = -\frac{1}{2} - \frac{4}{15} = \frac{-15 - 8}{30} = -\frac{23}{30}$$
$$\text{LHS} = x \times (y - z) = \frac{2}{3} \times \left(-\frac{3}{4} - \frac{2}{5}\right)$$ Take LCM of denominators $4$ and $5$, which is $20$:
$$-\frac{3}{4} - \frac{2}{5} = \frac{-15 - 8}{20} = -\frac{23}{20}$$ Multiply by $\frac{2}{3}$:
$$\text{LHS} = \frac{2}{3} \times \left(-\frac{23}{20}\right) = \frac{1 \times (-23)}{3 \times 10} = -\frac{23}{30}$$ Right Hand Side (RHS): $$\text{RHS} = (x \times y) - (x \times z) = \left(\frac{2}{3} \times -\frac{3}{4}\right) - \left(\frac{2}{3} \times \frac{2}{5}\right)$$ $$\left(\frac{2}{3} \times -\frac{3}{4}\right) = -\frac{2}{4} = -\frac{1}{2}$$ $$\left(\frac{2}{3} \times \frac{2}{5}\right) = \frac{4}{15}$$ $$\text{RHS} = -\frac{1}{2} - \frac{4}{15} = \frac{-15 - 8}{30} = -\frac{23}{30}$$
Since $\text{LHS} = -\frac{23}{30}$ and $\text{RHS} = -\frac{23}{30}$, therefore $\text{LHS} = \text{RHS}$. (Verified)
Review Exercise 1 Q4: Express in Ascending and Descending Order
(i) $\frac{5}{2}, \frac{7}{3}, \frac{9}{-4}, \frac{7}{2}$
Decimals: $2.5, 2.333, -2.25, 3.5$
• Ascending: $\frac{9}{-4} < \frac{7}{3} < \frac{5}{2} < \frac{7}{2}$
• Descending: $\frac{7}{2} > \frac{5}{2} > \frac{7}{3} > \frac{9}{-4}$
Decimals: $2.5, 2.333, -2.25, 3.5$
• Ascending: $\frac{9}{-4} < \frac{7}{3} < \frac{5}{2} < \frac{7}{2}$
• Descending: $\frac{7}{2} > \frac{5}{2} > \frac{7}{3} > \frac{9}{-4}$
(ii) $\frac{5}{3}, \frac{4}{5}, \frac{6}{8}, \frac{-3}{-2}$
Decimals: $1.667, 0.8, 0.75, 1.5$
• Ascending: $\frac{6}{8} < \frac{4}{5} < \frac{-3}{-2} < \frac{5}{3}$
• Descending: $\frac{5}{3} > \frac{-3}{-2} > \frac{4}{5} > \frac{6}{8}$
Decimals: $1.667, 0.8, 0.75, 1.5$
• Ascending: $\frac{6}{8} < \frac{4}{5} < \frac{-3}{-2} < \frac{5}{3}$
• Descending: $\frac{5}{3} > \frac{-3}{-2} > \frac{4}{5} > \frac{6}{8}$
(iii) $\frac{3}{4}, \frac{5}{3}, \frac{9}{-2}, \frac{7}{3}$
Decimals: $0.75, 1.667, -4.5, 2.333$
• Ascending: $\frac{9}{-2} < \frac{3}{4} < \frac{5}{3} < \frac{7}{3}$
• Descending: $\frac{7}{3} > \frac{5}{3} > \frac{3}{4} > \frac{9}{-2}$
Decimals: $0.75, 1.667, -4.5, 2.333$
• Ascending: $\frac{9}{-2} < \frac{3}{4} < \frac{5}{3} < \frac{7}{3}$
• Descending: $\frac{7}{3} > \frac{5}{3} > \frac{3}{4} > \frac{9}{-2}$
(iv) $\frac{5}{4}, \frac{2}{3}, \frac{6}{8}, \frac{-3}{2}$
Decimals: $1.25, 0.667, 0.75, -1.5$
• Ascending: $\frac{-3}{2} < \frac{2}{3} < \frac{6}{8} < \frac{5}{4}$
• Descending: $\frac{5}{4} > \frac{6}{8} > \frac{2}{3} > \frac{-3}{2}$
Decimals: $1.25, 0.667, 0.75, -1.5$
• Ascending: $\frac{-3}{2} < \frac{2}{3} < \frac{6}{8} < \frac{5}{4}$
• Descending: $\frac{5}{4} > \frac{6}{8} > \frac{2}{3} > \frac{-3}{2}$
Review Exercise 1 Q5: Fill in the Blanks with Proper Inequality Signs
(i) $a + \sqrt{5} > b + \sqrt{5} \implies$ $a$ $>$ $b$
(ii) $a > -\sqrt{7} \implies$ $3a$ $>$ $-3\sqrt{7}$
(iii) $\frac{1}{5} < \frac{1}{2} \implies$ $5$ $>$ $2$
(iv) $-\sqrt{11} > -11 \implies$ $-\frac{1}{\sqrt{11}}$ $<$ $-\frac{1}{11}$
(v) $3 < x \implies$ $-6$ $>$ $-2x$
Review Exercise 1 Q6: Find Absolute Value of $|3x - 4x + 1|$
Simplify the inner algebraic expression first: $3x - 4x + 1 = -x + 1 = 1 - x$.
The expression is $|1 - x|$.
When $x = 1$:
$|1 - (1)| = |0| = \mathbf{0}$
$|1 - (1)| = |0| = \mathbf{0}$
When $x = 2$:
$|1 - (2)| = |-1| = \mathbf{1}$
$|1 - (2)| = |-1| = \mathbf{1}$
When $x = 3$:
$|1 - (3)| = |-2| = \mathbf{2}$
$|1 - (3)| = |-2| = \mathbf{2}$
Frequently Asked Exam Questions & Concept Checks
1. Why is $\frac{22}{7}$ a rational number while $\pi$ is an irrational number?
$\frac{22}{7}$ is explicitly written as a ratio of two integers with non-zero denominator ($a=22, b=7$), and its decimal expansion is non-terminating but periodic ($3.\overline{142857}$). On the other hand, the true constant $\pi = \frac{\text{Circumference}}{\text{Diameter}}$ produces a non-terminating and non-recurring decimal ($3.14159265\dots$). Therefore, $\pi$ is irrational, and $\frac{22}{7}$ is merely an approximate rational fraction used for computation.
2. What happens to an inequality when multiplied or divided by a negative number?
The inequality direction is reversed. For instance, if $x < y$ and $c < 0$, then $cx > cy$. Example: $2 < 5 \implies 2 \times (-3) > 5 \times (-3) \implies -6 > -15$.
3. Can absolute value ever evaluate to a negative number?
No. Absolute value represents the physical geometric distance of a number from zero on the number line. Because distance is always a non-negative scalar quantity, $|x| \ge 0$ for all real numbers $x \in \mathbb{R}$.
4. Why does zero have no multiplicative inverse?
The multiplicative inverse of $a$ is defined as $\frac{1}{a}$ such that $a \times \frac{1}{a} = 1$. If $a = 0$, there is no real number $x$ such that $0 \times x = 1$ because zero multiplied by any real number is always $0$. Thus, division by zero is undefined.
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