Navigating through the newly revised CBSE Class 9 Mathematics curriculum (Ganita Manjari Part 1) requires a thorough foundational and algebraic understanding of the Real Number System, historical counting bases (base-10 decimal and ancient base-12 duodecimal systems), the set-theoretic hierarchy of numbers (Natural Numbers N, Whole Numbers W, Integers Z, Rational Numbers Q, and Irrational Numbers T/I), decimal expansion classifications (terminating, non-terminating recurring, and non-terminating non-recurring), the algebraic conversion of repeating decimals into p/q form, the geometric representation of square roots on the number line using the Pythagorean theorem, the construction of the Square Root Spiral (Spiral of Theodorus), algebraic operations on real numbers, the rationalisation of surd denominators using algebraic conjugates, and the laws of rational exponents for real powers. Chapter 3 of Class 9 Mathematics, “The World of Numbers”, deepens mathematical reasoning by proving that between any two rational numbers lie infinitely many rational and irrational numbers; demonstrates why numbers like √2, √3, and √5 cannot be expressed as ratios of integers; models ancient mercantile exchanges (such as Lothal barter calculations and Ishango bone prime sequences); and details the step-by-step arithmetic of rationalising complex surds. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and step-by-step responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from in-text “Think and Reflect” prompts and Exercise Sets 3.1, 3.2, 3.3, 3.4, and 3.5 to the complete End-of-Chapter Exercises (Questions 1 to 16 on Pages 58–62)—has been solved with exhaustive step-by-step derivations using strictly textbook-compliant methods. All mathematical calculations, decimal conversions, geometric constructions, and surd rationalisations follow a structured box format with clear ASCII number line diagrams using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Classification & Hierarchy of the Real Number System
================================================================================
THE REAL NUMBER SYSTEM HIERARCHY: R
================================================================================
REAL NUMBERS (R)
│
┌──────────────────────────────┴──────────────────────────────┐
▼ ▼
RATIONAL NUMBERS (Q) IRRATIONAL NUMBERS (T/I)
(Can be written as p/q, q ≠ 0) (Cannot be written as p/q)
• Terminating (e.g., 0.75, 5/2) • Non-Terminating & Non-Repeating
• Non-Terminating Recurring (e.g., 0.333...) • e.g., √2, √3, √5, π, 0.1010010001...
│
▼
INTEGERS (Z) = {..., -3, -2, -1, 0, 1, 2, 3, ...}
│
▼
WHOLE NUMBERS (W) = {0, 1, 2, 3, 4, ...}
│
▼
NATURAL NUMBERS (N) = {1, 2, 3, 4, 5, ...} (Counting Numbers)
================================================================================
| Number Category | Mathematical Definition | Standard Symbol | Decimal Expansion Behavior | Concrete Representative Examples |
|---|---|---|---|---|
| Natural Numbers | Positive counting numbers starting from 1. | N | Whole integers without decimal fraction. | 1, 2, 3, 4, 100, 525 |
| Whole Numbers | All natural numbers including Zero (0). | W | Whole integers without decimal fraction. | 0, 1, 2, 3, 4, 50 |
| Integers | All positive and negative whole numbers and 0. | Z | Whole integers without decimal fraction. | -15, -4, -1, 0, 3, 28 |
| Rational Numbers | Numbers expressible in the form p / q, where p, q are integers and q ≠ 0. | Q | Terminating OR Non-Terminating Recurring (Repeating). | 1/2 = 0.5, 2/3 = 0.666..., -7/4 = -1.75, 0 = 0/1 |
| Irrational Numbers | Numbers that cannot be expressed as p / q. | T / I | Non-Terminating and Non-Recurring (Non-Repeating). | √2 ≈ 1.4142..., √3 ≈ 1.7320..., √5 ≈ 2.2360..., π ≈ 3.1415... |
| Real Numbers | The complete union of all Rational and Irrational numbers (Q ∪ T). | R | Every point on the continuous number line represents a unique real number. | All rational and irrational numbers combined. |
2. Geometric Number Line Representation of Square Roots (Pythagoras Method)
================================================================================
GEOMETRIC PYTHAGOREAN CONSTRUCTION OF √2 ON NUMBER LINE
================================================================================
B (1 unit high)
│\
│ \ Hypotenuse OB = √(1² + 1²) = √2
1 unit │ \
│ \
───┼────────────┴────\───────●────────┼────────┼───► Number Line
-1 O A P 2 3
0 1 (√2 ≈ 1.414)
└──1──┘
1. Mark OA = 1 unit on the horizontal number line.
2. Draw perpendicular AB = 1 unit at point A.
3. Join OB: By Pythagoras Theorem, OB² = OA² + AB² = 1² + 1² = 2 ===> OB = √2.
4. With O as center and radius OB = √2, draw an arc intersecting the number line at P.
Point P represents the exact geometric location of √2 on the real number line.
================================================================================
3. Master Laws of Real Exponents & Surd Operations Sheet
| Law / Operation | Mathematical Rule | Example with Numerical Substitution |
|---|---|---|
| Product of Powers | aᵐ × aⁿ = aᵐ⁺ⁿ | 2^(2/3) × 2^(1/3) = 2^(2/3 + 1/3) = 2¹ = 2 |
| Quotient of Powers | aᵐ / aⁿ = aᵐ⁻ⁿ | 11^(1/2) / 11^(1/4) = 11^(1/2 - 1/4) = 11^(1/4) |
| Power of a Power | (aᵐ)ⁿ = aᵐˣⁿ | (3⁴)^(1/2) = 3^(4 × 1/2) = 3² = 9 |
| Product with Same Power | aᵐ × bᵐ = (a × b)ᵐ | 7^(1/2) × 8^(1/2) = (7 × 8)^(1/2) = 56^(1/2) = √56 |
| Negative Exponent | a⁻ⁿ = 1 / aⁿ | 5⁻² = 1 / 5² = 1 / 25 |
| Zero Exponent | a⁰ = 1 (where a ≠ 0) | (125)⁰ = 1 |
| Fractional Exponent (Radical) | a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ | 8^(2/3) = (³√8)² = (2)² = 4 |
| Conjugate Rationalisation | 1 / (√a ± √b) × (√a ∓ √b) / (√a ∓ √b) = (√a ∓ √b) / (a - b) | 1 / (√7 - √6) = (√7 + √6) / (7 - 6) = √7 + √6 |
NCERT In-Text Questions: “Think and Reflect”
Page No. 44: Think and Reflect (Questions 1 to 4)
Question 1 Is Zero (0) a rational number? Can you write it in the p/q form where p and q are integers and q is not equal to 0? [Exam Favorite]
Answer: Yes, Zero (0) is strictly a Rational Number.
- Mathematical Proof:
- Zero can be expressed in the standard p/q form as:
0 = 0 / 1 = 0 / 2 = 0 / 5 = 0 / (-3) - Here, the numerator p = 0 (which is an integer), and the denominator q = 1, 2, 5, … (which are non-zero integers, q ≠ 0).
- Since it satisfies all conditions of the definition, 0 is a rational number.
- Zero can be expressed in the standard p/q form as:
Question 2 Are all whole numbers natural numbers? Are all natural numbers whole numbers? Explain with reasons. [Exam Favorite]
Answer:
- (a) “All whole numbers are natural numbers” — FALSE.
- Reason: The number 0 is a whole number, but 0 is not a natural number (Natural numbers start from 1).
- (b) “All natural numbers are whole numbers” — TRUE.
- Reason: The set of natural numbers is N = {1, 2, 3, 4, …}. Every single natural number is contained within the set of whole numbers W = {0, 1, 2, 3, 4, …} (N is a subset of W).
Question 3 We know that Natural Numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are natural numbers closed under subtraction? Justify with counter-examples. [Exam Favorite]
Answer: No, Natural Numbers are NOT closed under subtraction.
- Mathematical Justification:
- Closure under subtraction requires that for any two natural numbers a and b, the difference (a – b) must always be a natural number.
- Example 1:
5 - 3 = 2(2 is a natural number). - Counter-Example 2:
3 - 5 = -2(-2 is a negative integer, NOT a natural number). - Counter-Example 3:
4 - 4 = 0(0 is a whole number, NOT a natural number).
- Since subtraction can produce negative integers or zero, natural numbers are not closed under subtraction.
Question 4 Ancient Indians used the joints of their fingers to count, a practice still observed today. Each of the four fingers has 3 joints (phalanges), and the thumb is used as a pointer to count them. How many can you count on one single hand? How does this practice relate to the ancient base-12 (duodecimal) counting system? [Exam Favorite]
Answer:
================================================================================
EXPLANATION OF ANCIENT BASE-12 FINGER-JOINT COUNTING:
--------------------------------------------------------------------------------
1. COUNTING CAPACITY ON ONE HAND:
• Number of counting fingers (excluding the thumb) = 4 (Index, Middle, Ring, Little).
• Number of visible joints/lines on each finger = 3 joints.
• Total count on one hand = 4 fingers × 3 joints per finger = 12.
2. RELATION TO THE BASE-12 (DUODECIMAL) SYSTEM:
• Because 12 could be counted on a single hand using the thumb as a pointer,
ancient civilizations (including ancient Indian and Mesopotamian traders)
adopted 12 as a fundamental base number.
• Using the 5 fingers of the second hand to track completed dozens allowed
counting up to 12 × 5 = 60 (origin of 60 seconds in a minute, 60 minutes in
an hour, and 360 degrees in a circle).
• This finger-joint arithmetic explains the historical origin of dozens (12 units),
gross (144 units), and 24 hours in a day.
================================================================================
NCERT Exercise Set 3.1 & 3.2: Historical Numbers & Barter Math (Pages 46–50)
Question 1 A merchant in the Harappan port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 copper ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with? [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP ARITHMETIC SOLUTION (LOTHAL BARTER RATIO): -------------------------------------------------------------------------------- GIVEN DATA: • Exchange Rate: 2 bags of spices = 15 copper ingots • Total bags of spices to exchange = 12 bags STEP 1: Find the number of 2-bag units Number of 2-bag units = 12 / 2 = 6 units STEP 2: Calculate Total Copper Ingots Total Copper Ingots = 6 units × 15 ingots per unit = 90 copper ingots Proportion Method: Let total ingots be x. 15 / 2 = x / 12 2 × x = 15 × 12 2x = 180 x = 180 / 2 = 90 FINAL ANSWER: The merchant will leave the Lothal market with 90 copper ingots. ================================================================================
Question 2 Look at the sequence of carved markings on one column of the ancient Ishango bone: 11, 13, 17, 19. (i) What mathematical property do these numbers have in common? (ii) List the next three numbers that fit this pattern. [Exam Favorite]
Answer:
- (i) Common Mathematical Property: The numbers 11, 13, 17, and 19 are all Prime Numbers (natural numbers greater than 1 that have exactly two distinct factors: 1 and the number itself).
- (ii) Next Three Numbers in the Sequence: The next three consecutive prime numbers after 19 are 23, 29, and 31.
Question 3 Perform the following arithmetic operations on rational numbers and express the final result in its lowest simplified terms: (a) (2 / 5) + (3 / 10) (b) (7 / 12) - (5 / 8) (c) (3 / 4) × (8 / 9) (d) (5 / 6) ÷ (15 / 4) [Exam Favorite]
Answer:
================================================================================
STEP-BY-STEP FRACTION OPERATIONS:
--------------------------------------------------------------------------------
(a) ADDITION: (2 / 5) + (3 / 10)
LCM of 5 and 10 = 10
= (2 × 2) / 10 + (3 × 1) / 10 = (4 + 3) / 10 = 7 / 10
(b) SUBTRACTION: (7 / 12) - (5 / 8)
LCM of 12 and 8 = 24
= (7 × 2) / 24 - (5 × 3) / 24 = (14 - 15) / 24 = -1 / 24
(c) MULTIPLICATION: (3 / 4) × (8 / 9)
= (3 × 8) / (4 × 9) = 24 / 36
Dividing numerator and denominator by 12:
= 2 / 3
(d) DIVISION: (5 / 6) ÷ (15 / 4)
Multiply by the reciprocal of the second fraction:
= (5 / 6) × (4 / 15) = (5 × 4) / (6 × 15) = 20 / 90
Dividing numerator and denominator by 10:
= 2 / 9
FINAL ANSWER SUMMARY:
• (a) 7/10
• (b) -1/24
• (c) 2/3
• (d) 2/9
================================================================================
NCERT Exercise Set 3.3 & 3.4: Decimals & p/q Conversions (Pages 53–56)
Question 1 Find five rational numbers strictly between 3/5 and 4/5 using the equivalent fraction method. [Exam Favorite]
Answer:
================================================================================ EQUIVALENT FRACTION METHOD (BETWEEN 3/5 AND 4/5): -------------------------------------------------------------------------------- To find 5 rational numbers, multiply numerator and denominator by (5 + 1) = 6: • First Number : 3 / 5 = (3 × 6) / (5 × 6) = 18 / 30 • Second Number : 4 / 5 = (4 × 6) / (5 × 6) = 24 / 30 Integers lying between numerators 18 and 24 are: 19, 20, 21, 22, 23. Therefore, five rational numbers between 3/5 and 4/5 are: 19/30, 20/30 (or 2/3), 21/30 (or 7/10), 22/30 (or 11/15), 23/30. FINAL ANSWER: 19/30, 2/3, 7/10, 11/15, and 23/30. ================================================================================
Question 2 Express the following recurring decimals in the rational p/q form (where p and q are integers and q ≠ 0): (i) 0.666... = 0.6_bar (ii) 0.4777... = 0.47_bar (iii) 0.001001... = 0.001_bar [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP ALGEBRAIC CONVERSION INTO p/q FORM: -------------------------------------------------------------------------------- (i) CONVERTING 0.666... : Let x = 0.6666... ...(Equation 1) Since 1 digit repeats, multiply Equation 1 by 10: 10x = 6.6666... ...(Equation 2) Subtract Equation 1 from Equation 2: 10x - x = (6.6666...) - (0.6666...) 9x = 6 x = 6 / 9 = 2 / 3 ===> 0.666... = 2 / 3 -------------------------------------------------------------------------------- (ii) CONVERTING 0.4777... : Let x = 0.47777... ...(Equation 1) Multiply Equation 1 by 10 (to bring non-repeating digit to left of decimal): 10x = 4.7777... ...(Equation 2) Multiply Equation 2 by 10 (since 1 digit repeats): 100x = 47.7777... ...(Equation 3) Subtract Equation 2 from Equation 3: 100x - 10x = (47.7777...) - (4.7777...) 90x = 43 x = 43 / 90 ===> 0.4777... = 43 / 90 -------------------------------------------------------------------------------- (iii) CONVERTING 0.001001001... : Let x = 0.001001001... ...(Equation 1) Since 3 digits repeat, multiply Equation 1 by 1000: 1000x = 1.001001001... ...(Equation 2) Subtract Equation 1 from Equation 2: 1000x - x = (1.001001...) - (0.001001...) 999x = 1 x = 1 / 999 ===> 0.001001... = 1 / 999 FINAL ANSWER SUMMARY: (i) 0.666... = 2 / 3 (ii) 0.4777... = 43 / 90 (iii) 0.001001... = 1 / 999 ================================================================================
NCERT End-of-Chapter Exercises (Pages 58–62, Questions 1 to 16)
Question 1 Classify the following numbers as rational or irrational with mathematical reasons: (a) √23 (b) √225 (c) 0.3796 (d) 7.478478... (e) 1.101001000100001... [Exam Favorite]
Answer:
- (a)
√23: IRRATIONAL. 23 is a prime number and not a perfect square; its square root is non-terminating and non-recurring. - (b)
√225 = 15 = 15/1: RATIONAL. 225 is a perfect square (15² = 225), expressible in p/q form. - (c)
0.3796 = 3796 / 10000: RATIONAL. It has a terminating decimal expansion. - (d)
7.478478...: RATIONAL. It has a non-terminating recurring (repeating) decimal expansion. - (e)
1.101001000100001...: IRRATIONAL. The number of zeroes increases progressively after each 1; the decimal expansion is non-terminating and non-recurring.
Question 2 Without actual division, determine which of the following fractions have terminating decimal expansions: (i) 13 / 3125 (ii) 17 / 8 (iii) 64 / 455 (iv) 29 / 343 [Exam Favorite]
Answer:
================================================================================
TERMINATING DECIMAL TEST (Denominator Form 2ᵐ × 5ⁿ):
--------------------------------------------------------------------------------
(i) 13 / 3125:
Prime factorisation of denominator 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵ = 2⁰ × 5⁵
Since denominator is strictly of the form 2ᵐ × 5ⁿ ===> TERMINATING.
(ii) 17 / 8:
Prime factorisation of denominator 8 = 2 × 2 × 2 = 2³ = 2³ × 5⁰
Since denominator is strictly of the form 2ᵐ × 5ⁿ ===> TERMINATING.
(iii) 64 / 455:
Prime factorisation of denominator 455 = 5 × 7 × 13
Contains prime factors 7 and 13 (other than 2 and 5) ===> NON-TERMINATING RECURRING.
(iv) 29 / 343:
Prime factorisation of denominator 343 = 7 × 7 × 7 = 7³
Does not match 2ᵐ × 5ⁿ ===> NON-TERMINATING RECURRING.
FINAL ANSWER:
Only fractions (i) 13/3125 and (ii) 17/8 have terminating decimal expansions.
================================================================================
Question 3 Locate and construct √5 on the number line using a step-by-step geometric Pythagorean method. [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP GEOMETRIC CONSTRUCTION OF √5: -------------------------------------------------------------------------------- MATHEMATICAL PRINCIPLE: By Pythagoras Theorem: (Hypotenuse)² = (Base)² + (Perpendicular)² (√5)² = 2² + 1² = 4 + 1 = 5 ===> Hypotenuse = √5 units CONSTRUCTION STEPS: 1. Draw a horizontal straight line and mark the origin O at 0. 2. Mark Point A on the number line at a distance of 2 units to the right of O (OA = 2 units). 3. At point A, construct a perpendicular line segment AB of unit length 1 (AB = 1 unit). 4. Join points O and B to form a right-angled triangle OAB right-angled at A. 5. In right triangle OAB: OB = √[ OA² + AB² ] = √[ 2² + 1² ] = √[ 4 + 1 ] = √5 units. 6. Taking O as center and radius equal to OB = √5, use a compass to draw an arc that intersects the number line at Point P. 7. Point P on the number line represents the exact location of √5 (≈ 2.236 units). ================================================================================
Question 4 Construct the first four stages of the Square Root Spiral (Spiral of Theodorus) on paper starting from OP₁ = 1 and P₁P₂ = 1. State the lengths of the successive hypotenuses. [Exam Favorite]
Answer:
================================================================================ SQUARE ROOT SPIRAL (SPIRAL OF THEODORUS): -------------------------------------------------------------------------------- 1. Stage 1 (Triangle OP₁P₂): • Base OP₁ = 1 unit, Perpendicular P₁P₂ = 1 unit (P₁P₂ perpendicular to OP₁). • Hypotenuse OP₂ = √(1² + 1²) = √2 units. 2. Stage 2 (Triangle OP₂P₃): • Base OP₂ = √2 units, Perpendicular P₂P₃ = 1 unit (P₂P₃ perpendicular to OP₂). • Hypotenuse OP₃ = √((√2)² + 1²) = √(2 + 1) = √3 units. 3. Stage 3 (Triangle OP₃P₄): • Base OP₃ = √3 units, Perpendicular P₃P₄ = 1 unit (P₃P₄ perpendicular to OP₃). • Hypotenuse OP₄ = √((√3)² + 1²) = √(3 + 1) = √4 = 2 units. 4. Stage 4 (Triangle OP₄P₅): • Base OP₄ = 2 units, Perpendicular P₄P₅ = 1 unit (P₄P₅ perpendicular to OP₄). • Hypotenuse OP₅ = √(2² + 1²) = √(4 + 1) = √5 units. SUCCESSIVE HYPOTENUSE LENGTHS: OP₂ = √2, OP₃ = √3, OP₄ = 2 (or √4), OP₅ = √5. ================================================================================
Question 5 Rationalise the denominator of each of the following expressions: (a) 1 / √7 (b) 1 / (√7 - √6) (c) 1 / (√5 + √2) (d) 1 / (7 + 3√2) [Exam Favorite]
Answer:
================================================================================
STEP-BY-STEP RATIONALISATION CALCULATIONS:
--------------------------------------------------------------------------------
(a) 1 / √7:
Multiply numerator and denominator by √7:
= (1 × √7) / (√7 × √7) = √7 / 7
(b) 1 / (√7 - √6):
Multiply numerator and denominator by conjugate (√7 + √6):
= [ 1 × (√7 + √6) ] / [ (√7 - √6)(√7 + √6) ]
Using identity (a - b)(a + b) = a² - b²:
= (√7 + √6) / [ (√7)² - (√6)² ]
= (√7 + √6) / (7 - 6) = (√7 + √6) / 1 = √7 + √6
(c) 1 / (√5 + √2):
Multiply numerator and denominator by conjugate (√5 - √2):
= [ 1 × (√5 - √2) ] / [ (√5 + √2)(√5 - √2) ]
= (√5 - √2) / [ (√5)² - (√2)² ]
= (√5 - √2) / (5 - 2) = (√5 - √2) / 3
(d) 1 / (7 + 3√2):
Multiply numerator and denominator by conjugate (7 - 3√2):
= [ 1 × (7 - 3√2) ] / [ (7 + 3√2)(7 - 3√2) ]
= (7 - 3√2) / [ (7)² - (3√2)² ]
Note: (3√2)² = 3² × (√2)² = 9 × 2 = 18
= (7 - 3√2) / (49 - 18) = (7 - 3√2) / 31
FINAL ANSWER SUMMARY:
• (a) √7 / 7
• (b) √7 + √6
• (c) (√5 - √2) / 3
• (d) (7 - 3√2) / 31
================================================================================
Question 6 Find the values of a and b if: (5 + 2√3) / (7 + 4√3) = a + b√3 [Exam Favorite]
Answer:
================================================================================
STEP-BY-STEP CONJUGATE MULTIPLICATION & EQUATING COEFFICIENTS:
--------------------------------------------------------------------------------
Given: (5 + 2√3) / (7 + 4√3) = a + b√3
STEP 1: Rationalise the Left-Hand Side (Multiply by conjugate 7 - 4√3)
LHS = [ (5 + 2√3)(7 - 4√3) ] / [ (7 + 4√3)(7 - 4√3) ]
STEP 2: Expand the Numerator
Numerator = 5·(7 - 4√3) + 2√3·(7 - 4√3)
= 35 - 20√3 + 14√3 - 8·(√3)²
= 35 - 6√3 - 8·(3)
= 35 - 6√3 - 24
= 11 - 6√3
STEP 3: Simplify the Denominator
Denominator = (7)² - (4√3)²
= 49 - (16 × 3) = 49 - 48 = 1
STEP 4: Equate with a + b√3
LHS = (11 - 6√3) / 1 = 11 - 6√3
11 - 6√3 = a + b√3
Equating rational and irrational parts:
• a = 11
• b = -6
FINAL ANSWER:
The values are a = 11 and b = -6.
================================================================================
Question 7 Simplify each of the following expressions using the laws of exponents: (a) 2^(2/3) × 2^(1/5) (b) (1 / 3³)^7 (c) 11^(1/2) / 11^(1/4) (d) 7^(1/2) × 8^(1/2) [Exam Favorite]
Answer:
================================================================================
EVALUATION USING LAWS OF EXPONENTS:
--------------------------------------------------------------------------------
(a) 2^(2/3) × 2^(1/5):
Using law aᵐ × aⁿ = aᵐ⁺ⁿ:
= 2^(2/3 + 1/5)
LCM of 3 and 5 = 15 ===> (2×5 + 1×3) / 15 = (10 + 3) / 15 = 13 / 15
= 2^(13/15)
(b) (1 / 3³)^7:
Using law (1 / aᵐ)ⁿ = (a⁻ᵐ)ⁿ = a⁻ᵐˣⁿ:
= (3⁻³)^7 = 3^(-3 × 7) = 3⁻²¹ = 1 / 3²¹
(c) 11^(1/2) / 11^(1/4):
Using law aᵐ / aⁿ = aᵐ⁻ⁿ:
= 11^(1/2 - 1/4)
(1/2 - 1/4) = (2 - 1) / 4 = 1 / 4
= 11^(1/4)
(d) 7^(1/2) × 8^(1/2):
Using law aᵐ × bᵐ = (a × b)ᵐ:
= (7 × 8)^(1/2) = 56^(1/2) = √56
FINAL ANSWER SUMMARY:
• (a) 2^(13/15)
• (b) 1 / 3²¹ (or 3⁻²¹)
• (c) 11^(1/4)
• (d) √56 (or 56^(1/2))
================================================================================
Question 8 Simplify the radical expression: (3 + √3)(2 + √2). [Exam Favorite]
Answer:
================================================================================
EXPANSION USING DISTRIBUTIVE PROPERTY:
--------------------------------------------------------------------------------
(3 + √3)(2 + √2) = 3·(2 + √2) + √3·(2 + √2)
= (3 × 2) + (3 × √2) + (2 × √3) + (√3 × √2)
= 6 + 3√2 + 2√3 + √6
FINAL ANSWER:
6 + 3√2 + 2√3 + √6
================================================================================
Question 9 Simplify: (√5 - √2)(√5 + √2). [Exam Favorite]
Answer:
================================================================================
USING ALGEBRAIC IDENTITY (a - b)(a + b) = a² - b²:
--------------------------------------------------------------------------------
(√5 - √2)(√5 + √2) = (√5)² - (√2)²
= 5 - 2 = 3
FINAL ANSWER:
The value is 3.
================================================================================
Question 10 If x = 2 + √3, find the value of x + (1 / x). [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP EVALUATION: -------------------------------------------------------------------------------- Given: x = 2 + √3 STEP 1: Find 1 / x by rationalising the denominator 1 / x = 1 / (2 + √3) Multiply by conjugate (2 - √3): 1 / x = [ 1 × (2 - √3) ] / [ (2 + √3)(2 - √3) ] 1 / x = (2 - √3) / [ (2)² - (√3)² ] 1 / x = (2 - √3) / (4 - 3) = (2 - √3) / 1 = 2 - √3 STEP 2: Calculate x + (1 / x) x + (1 / x) = (2 + √3) + (2 - √3) x + (1 / x) = 2 + 2 + √3 - √3 = 4 FINAL ANSWER: The value of x + (1 / x) is 4. ================================================================================
Question 11 Evaluate: (64)^(1/2) and (32)^(1/5) and (125)^(1/3). [Exam Favorite]
Answer:
================================================================================ EVALUATING ROOTS OF PERFECT POWERS: -------------------------------------------------------------------------------- 1. (64)^(1/2) = (8²)^(1/2) = 8^(2 × 1/2) = 8¹ = 8 2. (32)^(1/5) = (2⁵)^(1/5) = 2^(5 × 1/5) = 2¹ = 2 3. (125)^(1/3) = (5³)^(1/3) = 5^(3 × 1/3) = 5¹ = 5 FINAL ANSWER: • (64)^(1/2) = 8 • (32)^(1/5) = 2 • (125)^(1/3) = 5 ================================================================================
Question 12 Evaluate: (9)^(3/2) and (16)^(3/4) and (125)^(-1/3). [Exam Favorite]
Answer:
================================================================================ EVALUATING FRACTIONAL EXPONENTS: -------------------------------------------------------------------------------- 1. (9)^(3/2) = (3²)^(3/2) = 3^(2 × 3/2) = 3³ = 27 2. (16)^(3/4) = (2⁴)^(3/4) = 2^(4 × 3/4) = 2³ = 8 3. (125)^(-1/3) = (5³)^(-1/3) = 5^(3 × -1/3) = 5⁻¹ = 1 / 5 = 0.2 FINAL ANSWER: • (9)^(3/2) = 27 • (16)^(3/4) = 8 • (125)^(-1/3) = 1/5 (or 0.2) ================================================================================
Question 13 Find an irrational number strictly lying between the two rational numbers 1/7 and 2/7. [Exam Favorite]
Answer:
================================================================================ FINDING AN IRRATIONAL NUMBER BETWEEN 1/7 AND 2/7: -------------------------------------------------------------------------------- STEP 1: Find Decimal Expansions • 1 / 7 = 0.142857142857... • 2 / 7 = 0.285714285714... STEP 2: Construct a Non-Terminating Non-Repeating Decimal between 0.1428... and 0.2857... We choose any non-repeating, non-terminating pattern starting between 0.14 and 0.28: Example: 0.1501001000100001... Another Example: 0.201001000100001... FINAL ANSWER: 0.1501001000100001... is an irrational number lying strictly between 1/7 and 2/7. ================================================================================
Question 14 Show that 0.2353535... can be expressed in the form p/q, where p and q are integers and q ≠ 0. [Exam Favorite]
Answer:
================================================================================ ALGEBRAIC PROOF: -------------------------------------------------------------------------------- Let x = 0.2353535... ...(Equation 1) Multiply Equation 1 by 10 (to bring non-repeating digit 2 before decimal point): 10x = 2.353535... ...(Equation 2) Since 2 digits (35) repeat, multiply Equation 2 by 100: 1000x = 235.353535... ...(Equation 3) Subtract Equation 2 from Equation 3: 1000x - 10x = (235.353535...) - (2.353535...) 990x = 233 x = 233 / 990 FINAL ANSWER: 0.2353535... = 233 / 990 (which is in the standard p/q form). ================================================================================
Question 15 State whether the sum or product of two irrational numbers is always irrational. Justify with examples. [Exam Favorite]
Answer: No, the sum or product of two irrational numbers is NOT always irrational (it can be rational or irrational).
- Case 1: Sum of Two Irrationals:
- Irrationals √3 and -√3: Sum = √3 + (-√3) = 0 (Rational).
- Irrationals √2 and √3: Sum = √2 + √3 (Irrational).
- Case 2: Product of Two Irrationals:
- Irrationals √8 and √2: Product = √8 × √2 = √16 = 4 (Rational).
- Irrationals √2 and √3: Product = √2 × √3 = √6 (Irrational).
Question 16 Prove that (√3 - √2)² is an irrational number. [Exam Favorite]
Answer:
================================================================================
EXPANSION & IRRATIONALITY PROOF:
--------------------------------------------------------------------------------
Using algebraic identity (a - b)² = a² - 2ab + b²:
(√3 - √2)² = (√3)² - 2·(√3)(√2) + (√2)²
= 3 - 2√6 + 2
= 5 - 2√6
CONCLUSION:
• 5 is a Rational Number.
• 2√6 is an Irrational Number (since √6 is irrational).
• The difference between a rational number and an irrational number is
always strictly IRRATIONAL.
• Hence, 5 - 2√6 is strictly an IRRATIONAL NUMBER.
================================================================================
Frequently Asked Questions (FAQs) – Class 9 Maths Chapter 3
Question 1: What is the fundamental difference between a Rational and an Irrational number? [Exam Favorite] Answer:
- Rational Number: Can be expressed as p/q (q ≠ 0) with a decimal expansion that is terminating or non-terminating recurring.
- Irrational Number: Cannot be expressed as p/q (q ≠ 0) and has a decimal expansion that is non-terminating and non-recurring.
Question 2: Is π a rational or irrational number? What about 22/7? [Exam Favorite] Answer:
- π is strictly an IRRATIONAL number (its exact decimal value 3.14159265… is non-terminating and non-recurring).
- 22/7 is a RATIONAL number (p/q form), used only as an approximate working value for π.
Question 3: How many rational numbers exist between any two distinct rational numbers? [Exam Favorite] Answer: There are infinitely many rational numbers (and infinitely many irrational numbers) between any two distinct rational numbers (Dense property).
Question 4: What is the conjugate of the binomial surd 5 - 2√3? [Exam Favorite] Answer: The conjugate is 5 + 2√3.
Question 5: What is the value of (a + √b)(a - √b)? [Exam Favorite] Answer: Using (x + y)(x – y) = x² – y², the value is a² - b.
Question 6: Is the square root of every positive integer irrational? [Exam Favorite] Answer: No. Square roots of perfect square integers are rational numbers (e.g., √4 = 2, √9 = 3, √16 = 4, √25 = 5).
Question 7: Simplify (√11 - √7)(√11 + √7). [Exam Favorite] Answer: = (√11)² - (√7)² = 11 - 7 = 4.
Question 8: Write the simplest rationalising factor of √50. [Exam Favorite] Answer: √50 = √(25 × 2) = 5√2. The simplest rationalising factor is √2 (since 5√2 × √2 = 10).
Question 9: Find the value of (81)^(3/4). [Exam Favorite] Answer: (81)^(3/4) = (3⁴)^(3/4) = 3^(4 × 3/4) = 3³ = 27.
Question 10: Find the value of (256)^(-1/4). [Exam Favorite] Answer: (256)^(-1/4) = (4⁴)^(-1/4) = 4^(4 × -1/4) = 4⁻¹ = 1 / 4 (or 0.25).
Question 11: Under what condition on prime factorisation of q does p/q have a terminating decimal? [Exam Favorite] Answer: When the prime factorisation of the denominator q contains only powers of 2, powers of 5, or both (q = 2ᵐ × 5ⁿ, where m, n are whole numbers).
Question 12: What is the value of (√5 + √2)²? [Exam Favorite] Answer: = (√5)² + 2(√5)(√2) + (√2)² = 5 + 2√10 + 2 = 7 + 2√10.
Question 13: What is the product of a non-zero rational number and an irrational number? [Exam Favorite] Answer: The product is always strictly an IRRATIONAL number (e.g., 3 × √2 = 3√2).
Question 14: Convert 0.9999... into p/q form. [Exam Favorite] Answer: Let x = 0.999… ===> 10x = 9.999… ===> 9x = 9 ===> x = 9 / 9 = 1.
Question 15: Find the value of (32)^(-2/5). [Exam Favorite] Answer: (32)^(-2/5) = (2⁵)^(-2/5) = 2^(5 × -2/5) = 2⁻² = 1 / 2² = 1 / 4 (or 0.25).
Mastering the NCERT Solutions for Class 9 Mathematics Chapter 3 (Ganita Manjari Part 1), “The World of Numbers”, equips students with the number system hierarchy, repeating decimal conversions, surd rationalisations, geometric square root constructions, and exponential laws required for top performance in CBSE examinations. Review the step-by-step algebraic box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.