Chapter 1 — Number Systems | Class 9 CBSE Maths

Section 1.1

Types of Numbers

Every number belongs to a family. Here's the complete hierarchy from natural numbers to real numbers.

ℕ

Natural Numbers

Counting numbers starting from 1

1, 2, 3, 4, 5, ...

𝕎

Whole Numbers

Natural numbers + zero

0, 1, 2, 3, 4, ...

ℤ

Integers

Whole + all negatives. Z = "zahlen" (German: to count)

...-2,-1, 0, 1, 2...

ℚ

Rational Numbers

Can be written as p/q, q≠0. Q = "quotient"

½, -3/4, 0.75, -25

ℚ'

Irrational Numbers

Cannot be written as p/q. Non-terminating, non-recurring

√2, √3, π, e

ℝ

Real Numbers

All rationals + all irrationals = every point on number line

Any decimal number

Definition — Rational Number

A number r is called a rational number if it can be written in the form p/q, where p and q are integers and q ≠ 0. Examples: ½, -3, 0, 7/4, -2005/2006

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Why q ≠ 0? Division by zero is undefined. If q = 0, then p/q has no meaning. That's why we always insist q ≠ 0 in the definition of rational numbers.

Subset Relationships

ℕ ⊂ 𝕎 ⊂ ℤ ⊂ ℚ ⊂ ℝ

Every natural number is a whole number, every whole number is an integer, every integer is rational, every rational is real.

Common Misconceptions

❌ Every whole number is a natural number → FALSE (0 is whole but not natural)
❌ Every rational is an integer → FALSE (3/5 is not an integer)
✓ Every integer IS a rational (write as m/1)

Section 1.2

Irrational Numbers

Numbers that cannot be expressed as a ratio of integers — discovered by the ancient Greeks around 400 BC.

Definition — Irrational Number

A number s is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Its decimal expansion is non-terminating and non-recurring.

Famous Irrational Numbers

Number	Decimal Expansion	Why Irrational
√2	1.41421356237...	Non-terminating, non-recurring. Proven irrational by Pythagoreans (~400 BC)
√3	1.73205080757...	Proved by Theodorus of Cyrene (~425 BC)
√5	2.23606797749...	Proved by Theodorus of Cyrene
π	3.14159265358...	Proved irrational by Lambert and Legendre (late 1700s). Note: 22/7 ≈ π but π ≠ 22/7
0.10110111...	0.10110111011110...	Pattern of 1s keeps growing — never repeats a fixed block

Construction

Locating √2 and √3 on the Number Line

Draw a unit square OABC (each side = 1). By Pythagoras: OB = √(1² + 1²) = √2

Place O at zero on the number line. Draw arc with centre O, radius OB = √2. It hits the number line at P = √2 ≈ 1.414

Draw BD ⊥ OB with BD = 1 unit. By Pythagoras: OD = √((√2)² + 1²) = √3

Draw arc with centre O, radius OD = √3. It hits the number line at Q = √3 ≈ 1.732

💡

General rule: To locate √n, first locate √(n−1), then add a unit perpendicular and use Pythagoras. We can locate √n for any positive integer n this way!

Section 1.3

Real Numbers & Decimal Expansions

The type of decimal expansion tells us whether a number is rational or irrational.

Terminating Decimals

The remainder becomes zero at some stage. Decimal ends after finite steps.

7/8 = 0.875 ✓
1/2 = 0.5 ✓
639/250 = 2.556 ✓

Non-Terminating Recurring

Remainder repeats, giving a repeating block of digits. Written with a bar.

10/3 = 3.3̄ (block: 3)
1/7 = 0.̄142857 (block: 142857)
1.27̄ (block: 27)

Key Theorem — Decimal Classification

Rational: Decimal expansion is either terminating or non-terminating recurring.
Irrational: Decimal expansion is non-terminating and non-recurring.
These two properties together identify every real number as exactly one type.

Key Technique

Converting Recurring Decimals to p/q Form

The trick: multiply by 10ⁿ where n = number of repeating digits, then subtract to cancel the recurring part.

Example: 0.3̄ (1 digit repeating)

Let x = 0.3333...

10x = 3.3333... = 3 + x

9x = 3 → x = 1/3

Example: 1.2̄7 (2 digits repeating)

Let x = 1.272727...

100x = 127.2727... = 126 + x

99x = 126 → x = 14/11

🎯

Rule of thumb: If n digits repeat, multiply by 10ⁿ. If some digits don't repeat before the block, adjust accordingly (see 0.235̄ example below).

Section 1.4

Operations on Real Numbers

Rules for adding, subtracting, multiplying and dividing rational and irrational numbers.

Key Properties

(i) Rational ± Irrational = Irrational (e.g., 2 + √3 is irrational)
(ii) Non-zero Rational × Irrational = Irrational (e.g., 2√3 is irrational)
(iii) Irrational ± Irrational = May be rational or irrational (e.g., √6 + (−√6) = 0 which is rational)

Square Root Identities — The 6 Laws

Let a and b be positive real numbers. These identities are used constantly in simplification and rationalisation:

#	Identity	Example
(i)	√(ab) = √a · √b	√12 = √4·√3 = 2√3
(ii)	√(a/b) = √a / √b	√(9/4) = 3/2
(iii)	(√a+√b)(√a−√b) = a−b	(√5+√3)(√5−√3) = 5−3 = 2
(iv)	(a+√b)(a−√b) = a²−b	(3+√2)(3−√2) = 9−2 = 7
(v)	(√a+√b)(√c+√d) = √ac+√ad+√bc+√bd	(√2+√3)(√5+√7) = √10+√14+√15+√21
(vi)	(√a+√b)² = a + 2√(ab) + b	(√3+√7)² = 3+2√21+7 = 10+2√21

Rationalising the Denominator

When the denominator contains a surd (√), we multiply numerator and denominator by the conjugate to make the denominator rational.

Form	Multiply by	Result
1/√a	√a/√a	√a/a
1/(a+√b)	(a−√b)/(a−√b)	(a−√b)/(a²−b)
1/(√a+√b)	(√a−√b)/(√a−√b)	(√a−√b)/(a−b)
1/(√a−√b)	(√a+√b)/(√a+√b)	(√a+√b)/(a−b)

Geometric Construction

Representing √x on the Number Line (for any positive real x)

Proof: OC = OD = OA = (x+1)/2 (radii). OB = (x−1)/2. By Pythagoras: BD² = OD² − OB² = ((x+1)/2)² − ((x−1)/2)² = x. So BD = √x. ✓

Section 1.5

Laws of Exponents for Real Numbers

Extending the laws of exponents from natural numbers to rational and real exponents.

Key Definition — nth Root

Let a > 0 be a real number and n a positive integer. Then ⁿ√a = b if bⁿ = a and b > 0.
In exponent notation: ⁿ√a = a^(1/n) and a^(m/n) = (ⁿ√a)ᵐ = ⁿ√(aᵐ)

Laws of Exponents (Extended to Rational Exponents)

Let a > 0 be a real number and p, q be rational numbers:

Law	Formula	Example
Product Rule	aᵖ · aᵍ = a^(p+q)	2^(2/3) · 2^(1/3) = 2^1 = 2
Power of Power	(aᵖ)ᵍ = a^(pq)	(3^(1/5))⁴ = 3^(4/5)
Quotient Rule	aᵖ / aᵍ = a^(p−q)	7^(1/5) / 7^(1/3) = 7^(−2/15)
Product Base	aᵖ · bᵖ = (ab)ᵖ	13^(1/5) · 17^(1/5) = 221^(1/5)
Zero Exponent	a⁰ = 1	5⁰ = 1, (√7)⁰ = 1
Negative Exponent	a^(−n) = 1/aⁿ	2^(−3) = 1/8

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Key insight: a^(m/n) = (a^(1/n))^m = (ⁿ√a)^m. You can either take the root first then raise to power, or raise to power first then take the root. Both give the same answer!

Textbook Examples

All Worked Examples — Step by Step

Every example from the NCERT textbook, fully solved with justification at each step.

Example 1

True or False: (i) Every whole number is a natural number. (ii) Every integer is a rational number. (iii) Every rational number is an integer.

FALSE. Zero (0) is a whole number but not a natural number. ℕ = {1,2,3,...}, so 0 ∉ ℕ.

TRUE. Every integer m can be written as m/1, which is of the form p/q with q=1 ≠ 0. So all integers are rational.

iii

FALSE. 3/5 is a rational number but not an integer (it lies between 0 and 1, and is not a whole number).

Example 2

Find five rational numbers between 1 and 2.

Method 1

Use midpoint method: midpoint of 1 and 2 is 3/2. Then find midpoints of (1, 3/2) and (3/2, 2) etc.
Numbers: 3/2, 5/4, 7/4, 11/8, 13/8

Method 2

Write 1 and 2 with denominator 6 (= 5+1): 1 = 6/6 and 2 = 12/6.
So: 7/6, 8/6, 9/6, 10/6, 11/6 — simplified: 7/6, 4/3, 3/2, 5/3, 11/6

Five rational numbers: 7/6, 4/3, 3/2, 5/3, 11/6 (infinitely many more exist!)

📌

Important: There are infinitely many rational numbers between any two given rational numbers. We just pick any five.

Example 5

Find decimal expansions of 10/3, 7/8 and 1/7.

10/3 (non-terminating recurring)

10 ÷ 3 = 3.333...
Remainder: 1, 1, 1, 1... (repeats)
= 3.3̄

7/8 (terminating)

7 ÷ 8 = 0.875
Remainders: 6, 4, 0
Remainder becomes 0 → terminates

1/7 (non-terminating recurring)

1 ÷ 7 = 0.142857142857...
Remainders: 3,2,6,4,5,1,3,2,6,4,5,1...
Block of 6 digits repeats
= 0.1̄42857̄

Note: # repeating digits < divisor (7). Here 6 < 7 ✓

Examples 6–9

Expressing terminating and recurring decimals as p/q

Ex 6: Show 3.142678 is rational

Terminating decimal → multiply out: 3.142678 = 3142678/1000000

= 3142678/1000000 (rational ✓)

Ex 7: Show 0.3̄ = p/q

Let x = 0.3333...

10x = 3.333... = 3 + x

9x = 3 → x = 1/3

0.3̄ = 1/3

Ex 8: Show 1.27̄ = p/q

x = 1.272727... (2 digits repeat)

100x = 127.2727... = 126 + x

99x = 126 → x = 126/99 = 14/11

1.27̄ = 14/11

Ex 9: Show 0.235̄ = p/q

x = 0.2353535... (2 digits repeat from 3rd place)

100x = 23.535... = 23.3 + x

99x = 23.3 = 233/10 → x = 233/990

0.235̄ = 233/990

Example 10

Find an irrational number between 1/7 and 2/7.

1/7 = 0.142857̄ and 2/7 = 0.285714̄

We need a number between these that is non-terminating and non-recurring

Choose a number in that range with no repeating pattern: 0.150150015000150000... (pattern of 1s grows)

0.150150015000150000... is irrational and lies between 1/7 and 2/7

Example 11

Check whether 7√5, 7/√5, √2 + 21, π − 2 are irrational.

√5 = 2.236...(irrational) → 7√5 = 15.652... (non-terminating non-recurring) → Irrational ✓

7/√5 = 7√5/5 = 3.1304... (non-terminating non-recurring) → Irrational ✓

√2 + 21 = 1.414... + 21 = 22.414... (non-terminating non-recurring) → Irrational ✓

π − 2 = 3.1415... − 2 = 1.1415... (non-terminating non-recurring) → Irrational ✓

Examples 12–14

Operations with surds

Ex 12

Add (2√2+5√3) + (√2−3√3)
= (2√2+√2) + (5√3−3√3) = 3√2 + 2√3

Ex 13

Multiply 6√5 × 2√5
= 6×2×√5×√5 = 12×5 = 60

Ex 14

Divide 8√15 ÷ 2√3
= (8/2) × (√15/√3) = 4 × √(15/3) = 4×√5 = 4√5

Example 15

Simplify: (i) (5+√7)(2+√5) (ii) (5+√5)(5−√5) (iii) (√3+√7)² (iv) (√11−√7)(√11+√7)

(5+√7)(2+√5) = 10 + 5√5 + 2√7 + √35 Identity (v): expand using FOIL

(5+√5)(5−√5) = 5² − (√5)² = 25 − 5 = 20 Identity (iv): (a+√b)(a−√b) = a²−b

iii

(√3+√7)² = 3 + 2√(3×7) + 7 = 10 + 2√21 Identity (vi): (√a+√b)² = a+2√ab+b

(√11−√7)(√11+√7) = 11 − 7 = 4 Identity (iii): (√a+√b)(√a−√b) = a−b

Examples 16–19

Rationalising the Denominator

Ex 16

Rationalise 1/√2
Multiply by √2/√2: 1/√2 = √2/2

Ex 17

Rationalise 1/(2+√3)
Multiply by (2−√3)/(2−√3): = (2−√3)/(4−3) = 2−√3

Ex 18

Rationalise 5/(√3−√5)
Multiply by (√3+√5)/(√3+√5): = 5(√3+√5)/(3−5) = (−5/2)(√3+√5)

Ex 19

Rationalise 1/(7+3√2)
Multiply by (7−3√2)/(7−3√2): = (7−3√2)/(49−18) = (7−3√2)/31

Example 20

Simplify using laws of exponents: (i) 2^(2/3)·2^(1/3) (ii) (3^(1/5))⁴ (iii) 7^(1/5)÷7^(1/3) (iv) 13^(1/5)·17^(1/5)

2^(2/3) · 2^(1/3) = 2^(2/3+1/3) = 2^(3/3) = 2¹ = 2 aᵖ·aᵍ = a^(p+q)

(3^(1/5))⁴ = 3^(4/5) (aᵖ)ᵍ = a^(pq)

iii

7^(1/5) / 7^(1/3) = 7^(1/5−1/3) = 7^(3/15−5/15) = 7^(−2/15) aᵖ/aᵍ = a^(p−q)

13^(1/5) · 17^(1/5) = (13×17)^(1/5) = 221^(1/5) aᵖ·bᵖ = (ab)ᵖ

NCERT Exercises

All Exercises — Fully Solved

Exercises 1.1 through 1.5, every question solved with clear steps.

Exercise 1.1

Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Yes, zero is a rational number.

0 = 0/1, 0/2, 0/3, ... — all have the form p/q where p = 0 and q is any non-zero integer.

Yes, 0 is rational: 0 = 0/1 (p=0, q=1, q≠0) ✓

Find six rational numbers between 3 and 4.

Write 3 and 4 with denominator 7 (= 6+1): 3 = 21/7, 4 = 28/7

Numbers between: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7

22/7, 23/7, 24/7, 25/7, 26/7, 27/7

Find five rational numbers between 3/5 and 4/5.

Convert to denominator 30 (= 5×6): 3/5 = 18/30, 4/5 = 24/30

Five numbers between: 19/30, 20/30, 21/30, 22/30, 23/30

19/30, 2/3, 7/10, 11/15, 23/30

State whether true or false: (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number.

TRUE. ℕ = {1,2,3,...} ⊂ 𝕎 = {0,1,2,3,...}. Every natural number is also a whole number.

FALSE. −3 is an integer but not a whole number. Whole numbers are non-negative.

iii

FALSE. 1/2 is a rational number but not a whole number.

(i) True (ii) False (iii) False

Exercise 1.2

State whether true or false: (i) Every irrational number is a real number. (ii) Every point on number line is of form √m. (iii) Every real number is an irrational number.

TRUE. Real numbers = rationals ∪ irrationals. All irrationals are real numbers by definition.

FALSE. Points like 2, 3, −1/2 are on the number line but 2 = √4, yes, but −1/2 cannot be written as √m for any natural number m (√m ≥ 0 always).

iii

FALSE. Real numbers include both rationals (like 1/2, 3) and irrationals. Not every real number is irrational.

(i) True (ii) False (iii) False

Are the square roots of all positive integers irrational? Give an example of square root that is rational.

No. Square roots of perfect squares are rational.

√4 = 2 (rational), √9 = 3 (rational), √16 = 4 (rational), √25 = 5 (rational)

No. Example: √4 = 2, which is rational (and an integer).

Show how √5 can be represented on the number line.

Take OA = 2 units on the number line with O at 0.

Construct AB = 1 unit perpendicular to OA at A.

OB = √(OA² + AB²) = √(4+1) = √5 (by Pythagoras)

Draw arc with centre O, radius OB = √5. It intersects the number line at P. P represents √5.

Use a right triangle with legs 2 and 1. Hypotenuse = √5. Swing arc to number line.

Exercise 1.3

Write the following in decimal form and name the type: (i) 36/100 (ii) 1/11 (iii) 4⅛ (iv) 3/13 (v) 2/11 (vi) 329/400

Fraction	Decimal	Type
(i) 36/100	0.36	Terminating
(ii) 1/11	0.09̄0̄9̄ = 0.0̄9̄	Non-terminating recurring
(iii) 4⅛ = 33/8	4.125	Terminating
(iv) 3/13	0.230769̄	Non-terminating recurring
(v) 2/11	0.18̄	Non-terminating recurring
(vi) 329/400	0.8225	Terminating

Express in p/q form: (i) 0.6̄ (ii) 0.4̄7̄ (iii) 0.00̄1̄

x = 0.666... → 10x = 6.666... = 6+x → 9x = 6 → x = 2/3

x = 0.4777... → 10x = 4.777..., 100x = 47.777... → 100x−10x = 43 → x = 43/90

iii

x = 0.001001... → 1000x = 1.001001... = 1+x → 999x = 1 → x = 1/999

(i) 2/3 (ii) 43/90 (iii) 1/999

Express 0.99999... in the form p/q. Are you surprised?

Let x = 0.9999...

10x = 9.9999... = 9 + x → 9x = 9 → x = 1

0.9999... = 1 ← Yes, surprising! As the 9s go on forever, the number equals exactly 1.

Classify as rational or irrational: (i) √23 (ii) √225 (iii) 0.3796 (iv) 7.478478... (v) 1.101001000100001...

Number	Classification	Reason
(i) √23	Irrational	23 is not a perfect square; decimal is non-terminating non-recurring
(ii) √225	Rational	√225 = 15, a natural number
(iii) 0.3796	Rational	Terminating decimal → rational
(iv) 7.478478...	Rational	Non-terminating recurring (block 478 repeats)
(v) 1.101001000...	Irrational	Non-terminating non-recurring (1s increase by 1 zero each time)

Exercise 1.4

Classify as rational or irrational: (i) 2−√5 (ii) (3+√23)−√23 (iii) 2√7/(7√7) (iv) 1/√2 (v) 2π

Number	Simplified	Type
(i) 2−√5	2 − 2.236... = −0.236...	Irrational (rational − irrational)
(ii) (3+√23)−√23	= 3	Rational (the √23 cancels!)
(iii) 2√7/(7√7)	= 2/7	Rational (√7 cancels)
(iv) 1/√2	= √2/2 = 0.707...	Irrational
(v) 2π	= 6.2831...	Irrational (rational × irrational)

Simplify: (i) (3+√3)(2+√2) (ii) (3+√3)(3−√3) (iii) (√5+√2)² (iv) (√5−√2)(√5+√2)

(3+√3)(2+√2) = 6+3√2+2√3+√6 Expand using FOIL

(3+√3)(3−√3) = 3²−(√3)² = 9−3 = 6 Identity (iv)

iii

(√5+√2)² = 5+2√10+2 = 7+2√10 Identity (vi)

(√5−√2)(√5+√2) = 5−2 = 3 Identity (iii)

(i) 6+3√2+2√3+√6 (ii) 6 (iii) 7+2√10 (iv) 3

Rationalise the denominators: (i) 1/√7 (ii) 1/(√7−√6) (iii) 1/(√5+√2) (iv) 1/(√7−2)

1/√7 × √7/√7 = √7/7

1/(√7−√6) × (√7+√6)/(√7+√6) = (√7+√6)/(7−6) = √7+√6

iii

1/(√5+√2) × (√5−√2)/(√5−√2) = (√5−√2)/(5−2) = (√5−√2)/3

1/(√7−2) × (√7+2)/(√7+2) = (√7+2)/(7−4) = (√7+2)/3

(i) √7/7 (ii) √7+√6 (iii) (√5−√2)/3 (iv) (√7+2)/3

Exercise 1.5

Find: (i) 64^(1/2) (ii) 32^(1/5) (iii) 125^(1/3)

64^(1/2) = √64 = 8

32^(1/5) = ⁵√32 = ⁵√(2⁵) = 2

iii

125^(1/3) = ³√125 = ³√(5³) = 5

(i) 8 (ii) 2 (iii) 5

Find: (i) 9^(3/2) (ii) 32^(2/5) (iii) 16^(3/4) (iv) 125^(−1/3)

9^(3/2) = (9^(1/2))³ = 3³ = 27

32^(2/5) = (32^(1/5))² = 2² = 4

iii

16^(3/4) = (16^(1/4))³ = 2³ = 8

125^(−1/3) = 1/125^(1/3) = 1/5

(i) 27 (ii) 4 (iii) 8 (iv) 1/5

Simplify: (i) 2^(2/3)·2^(1/5) (ii) (1/3³)⁷ (iii) 11^(1/2)÷11^(1/4) (iv) 7^(1/2)·8^(1/2)

2^(2/3)·2^(1/5) = 2^(2/3+1/5) = 2^(10/15+3/15) = 2^(13/15)

(1/3³)⁷ = (3⁻³)⁷ = 3⁻²¹ = 1/3²¹

iii

11^(1/2)/11^(1/4) = 11^(1/2−1/4) = 11^(1/4)

7^(1/2)·8^(1/2) = (7×8)^(1/2) = 56^(1/2) = √56 = 2√14

(i) 2^(13/15) (ii) 1/3²¹ (iii) 11^(1/4) (iv) 2√14

Study Strategy

10 Tips for CBSE Class 9 Students

Master Number Systems the smart way — these tips cover the most common mistakes and exam patterns.

Know the Hierarchy

ℕ ⊂ 𝕎 ⊂ ℤ ⊂ ℚ ⊂ ℝ. Every natural number is a whole number, every whole number is an integer, every integer is rational. Only 0 separates ℕ from 𝕎.

Decimal Type → Number Type

Terminating → Rational. Non-terminating recurring → Rational. Non-terminating non-recurring → Irrational. Memorise this table!

Perfect Squares ≠ Irrational

√4=2, √9=3, √16=4, √25=5 are ALL rational. Only non-perfect-square roots are irrational. Check: is the number a perfect square?

Recurring Decimal Trick

For n repeating digits, multiply by 10ⁿ. Subtract original. Solve. E.g., 2 digits repeat → multiply by 100, subtract, divide by 99.

Rationalisation Conjugate

Always multiply by conjugate: if denominator is (a+√b), multiply by (a−√b). If (√a+√b), multiply by (√a−√b). The key identity: (x+y)(x−y) = x²−y².

22/7 ≠ π !

π is irrational. 22/7 is just an approximation. The actual value of π = 3.14159265... which is non-terminating non-recurring. In exams, clarify this difference.

Laws of Exponents: Add Fractions

When multiplying same base: add exponents. When dividing: subtract. Convert fractions to same denominator first. E.g., 2/3 + 1/5 = 10/15 + 3/15 = 13/15.

Irrational × Irrational Can Be Rational

√2 × √2 = 2 (rational!). √6 × √6 = 6. Don't assume irrational × irrational = irrational. The result can go either way.

Use the (√a+√b)(√a−√b) Identity

This is the most-used identity in this chapter. It eliminates the surd and gives a rational number: (√5+√3)(√5−√3) = 5−3 = 2.

a^(m/n) — Two Ways

a^(m/n) = (ⁿ√a)^m = ⁿ√(a^m). Use whichever is easier. For 9^(3/2): first take √9=3, then cube → 27. Much easier than computing 9³=729 first!

Quick Reference

Chapter 1 — Formula & Fact Sheet

Concept	Formula / Rule	Example
Rational number	p/q, q≠0	½, -3/4, 0, 22
Terminating decimal	Denominator = 2ᵐ·5ⁿ	7/8 = 0.875 (2³)
n repeating digits → p/q	Multiply by 10ⁿ, subtract, solve	0.3̄: x=1/3
Product of square roots	√a·√b = √(ab)	√2·√8 = √16 = 4
Conjugate identity	(√a+√b)(√a−√b) = a−b	(√7+√5)(√7−√5) = 2
Rationalise 1/√a	= √a/a	1/√3 = √3/3
Rationalise 1/(a+√b)	= (a−√b)/(a²−b)	1/(2+√3) = 2−√3
nth root	ⁿ√a = a^(1/n)	⁵√32 = 32^(1/5) = 2
Fractional exponent	a^(m/n) = (ⁿ√a)^m	9^(3/2) = (√9)³ = 27
Product rule (exponents)	aᵖ·aᵍ = a^(p+q)	2^(1/3)·2^(2/3) = 2
Quotient rule	aᵖ/aᵍ = a^(p−q)	5^(3/4)/5^(1/4) = 5^(1/2)
Power of power	(aᵖ)ᵍ = a^(pq)	(4^(1/2))³ = 4^(3/2) = 8

CBSE Pattern Practice

50 Practice Questions

All CBSE question types — MCQ, 1-mark, 2-mark, 3-mark, and 5-mark questions with complete answers.

Section A — MCQ (1 mark each)

Which of the following is NOT a rational number?

MCQ

(a) √4

(b) 0.666...

(d) 22/7

Answer: (c) √5
√4 = 2 (rational), 0.666... = 2/3 (rational), 22/7 is a fraction (rational). √5 is irrational (5 is not a perfect square).

The decimal expansion of 1/7 is:

MCQ

(a) Terminating

(b) Non-term. non-recurring

(d) None

Answer: (c) Non-terminating recurring
1/7 = 0.142857142857... — the block 142857 repeats indefinitely.

Between any two rational numbers, there are:

MCQ

(a) Exactly one rational

(b) No rationals

(d) Finite rationals

Answer: (c) Infinitely many rationals
Between any two rational numbers r and s, the number (r+s)/2 is always rational, and we can repeat this process infinitely.

9^(3/2) equals:

MCQ

(a) 3

(b) 9

(d) 81

Answer: (c) 27
9^(3/2) = (9^(1/2))³ = 3³ = 27

The rationalising factor of 1/(2+√3) is:

MCQ

(a) 2+√3

(b) 2−√3

(d) 1/(2−√3)

Answer: (b) 2−√3
Multiply by the conjugate (2−√3): (2+√3)(2−√3) = 4−3 = 1 (rational). So 2−√3 is the rationalising factor.

(√5+√3)(√5−√3) equals:

MCQ

(a) 8

(b) √2

(d) √8

Answer: (c) 2
(√a+√b)(√a−√b) = a−b → (√5)²−(√3)² = 5−3 = 2

Which set satisfies ℕ ⊂ ? ⊂ ℚ directly?

MCQ

(a) ℝ

(b) ℤ

(d) Complex

Answer: (b) ℤ (Integers)
The chain is ℕ ⊂ 𝕎 ⊂ ℤ ⊂ ℚ. Integers (ℤ) come directly between natural numbers and rationals in this hierarchy.

2^(1/3) × 2^(2/3) equals:

MCQ

(a) 4

(b) 2^(1/9)

(d) 8

Answer: (c) 2
2^(1/3) × 2^(2/3) = 2^(1/3+2/3) = 2^(3/3) = 2^1 = 2

Which of the following is an irrational number?

MCQ

(a) 0.272727...

(b) 1.4142135...

(d) √9

Answer: (b) 1.4142135...
This is √2, which is non-terminating non-recurring → irrational. 0.272727... = 0.27̄ is recurring → rational. 3/7 is rational. √9=3 is rational.

If x = 0.999..., then x equals:

MCQ

(a) Less than 1

(b) Greater than 1

(d) Irrational

Answer: (c) Exactly 1
Let x = 0.999..., then 10x = 9.999... = 9+x → 9x = 9 → x = 1

Section B — Very Short Answer (1 mark each)

Write two irrational numbers between 1 and 2.

1 Mark

Answer: √2 ≈ 1.414... and √3 ≈ 1.732... are both irrational and lie between 1 and 2. (Many other answers are valid.)

Express 0.4̄ in the form p/q.

1 Mark

Answer: 4/9
x = 0.444..., 10x = 4.444... = 4+x, 9x = 4, x = 4/9

Evaluate: 27^(1/3)

1 Mark

Answer: 3
27^(1/3) = ³√27 = ³√(3³) = 3

Is √2 rational or irrational? Give one reason.

1 Mark

Answer: Irrational. √2 = 1.41421356... which is non-terminating and non-recurring. It cannot be expressed as p/q where p, q are integers.

Simplify: (√3+√7)(√3−√7)

1 Mark

Answer: −4
(√3)²−(√7)² = 3−7 = −4

Find a rational number between 3/7 and 4/7.

1 Mark

Answer: Midpoint = (3/7+4/7)/2 = (7/7)/2 = 1/2. So 1/2 (= 0.5) lies between 3/7 ≈ 0.428 and 4/7 ≈ 0.571. ✓

Write the decimal expansion of 2/11.

1 Mark

Answer: 0.18̄ = 0.181818...
Non-terminating recurring (block "18" repeats).

Simplify: (√5)²

1 Mark

Answer: 5
By definition, (√a)² = a for a > 0. So (√5)² = 5.

Is π rational? What is 22/7?

1 Mark

Answer: π is irrational. Its decimal expansion is non-terminating non-recurring (3.14159265...). 22/7 ≈ 3.142857... is only an approximation of π. π ≠ 22/7.

Evaluate: 125^(−1/3)

1 Mark

Answer: 1/5
125^(−1/3) = 1/125^(1/3) = 1/∛125 = 1/5

Section C — Short Answer (2 marks each)

Find four rational numbers between 1/3 and 1/2.

2 Mark

Answer:
Convert to common denominator 12: 1/3 = 4/12, 1/2 = 6/12.
Multiply by 5 more: 1/3 = 20/60, 1/2 = 30/60.
Four rationals: 21/60, 23/60, 25/60, 27/60 = 7/20, 23/60, 5/12, 9/20

Show that 0.235̄ = 233/990.

2 Mark

Answer:
Let x = 0.23535...
100x = 23.535... = 23.3 + x (the "35" block shifts)
Wait: 100x = 23.5353... = 23.3 + 0.2353... = 23.3 + x
99x = 23.3 = 233/10 → x = 233/990 ✓

Rationalise: 1/(√5+√2)

2 Mark

Answer:
Multiply by (√5−√2)/(√5−√2):
= (√5−√2)/((√5)²−(√2)²) = (√5−√2)/(5−2) = (√5−√2)/3

Simplify: (√3+√7)² − (√7−√3)²

2 Mark

Answer: 4√21
(√3+√7)² = 3+2√21+7 = 10+2√21
(√7−√3)² = 7−2√21+3 = 10−2√21
Difference = (10+2√21) − (10−2√21) = 4√21

Represent √5 on the number line.

2 Mark

Answer:
1. Mark O at 0 and A at 2 on number line (OA = 2).
2. Draw AB ⊥ OA with AB = 1.
3. OB = √(4+1) = √5 (Pythagoras).
4. Draw arc centre O, radius OB. It cuts number line at P = √5 ≈ 2.236.

Simplify: 2^(2/3) × 2^(1/5)

2 Mark

Answer: 2^(13/15)
2/3 + 1/5 = 10/15 + 3/15 = 13/15
2^(2/3) × 2^(1/5) = 2^(13/15)

Is 0.10110111011110... rational or irrational? Justify.

2 Mark

Answer: Irrational.
The decimal 0.10110111011110... is non-terminating (it goes on forever). The blocks are 1, 11, 111, 1111,... each adding a 1 — so no block of digits repeats. Therefore it is non-recurring. A non-terminating non-recurring decimal is irrational.

Simplify: (5+√2)(5−√2) + (√3+√7)²

2 Mark

Answer: 20 + 2√21
(5+√2)(5−√2) = 25−2 = 23
(√3+√7)² = 3+2√21+7 = 10+2√21
Sum = 23 + 10 + 2√21 = 33 + 2√21

Rationalise: 5/(3−√5)

2 Mark

Answer: 5(3+√5)/4
Multiply by (3+√5)/(3+√5):
= 5(3+√5)/(9−5) = 5(3+√5)/4

If a = 2+√3, find a − 1/a.

2 Mark

Answer: 2√3
1/a = 1/(2+√3) = (2−√3)/((2+√3)(2−√3)) = (2−√3)/(4−3) = 2−√3
a − 1/a = (2+√3)−(2−√3) = 2√3

Section D — Short Answer II (3 marks each)

Show that 1.272727... can be expressed as p/q where p = 14 and q = 11.

3 Mark

Answer:
Let x = 1.272727...
100x = 127.2727...
100x − x = 127.2727... − 1.2727...
99x = 126
x = 126/99 = 14/11
Verification: 14/11 = 1.272727... ✓

If p = 3+2√2 and q = 3−2√2, find p² + q².

3 Mark

Answer: 34
p+q = 6, pq = (3+2√2)(3−2√2) = 9−8 = 1
p² + q² = (p+q)² − 2pq = 36 − 2 = 34

Simplify: (3−√5)(3+√5) + (4+√7)(4−√7)

3 Mark

Answer: 13
(3−√5)(3+√5) = 9−5 = 4
(4+√7)(4−√7) = 16−7 = 9
Sum = 4 + 9 = 13

Rationalise the denominator of 1/(7+3√2) and simplify.

3 Mark

Answer: (7−3√2)/31
Multiply by (7−3√2)/(7−3√2):
Denominator: (7+3√2)(7−3√2) = 49−18 = 31
Result: (7−3√2)/31

Find the value of a and b if (3+√7)/(3−√7) = a + b√7.

3 Mark

Answer: a = 8, b = 3
Rationalise: (3+√7)/(3−√7) × (3+√7)/(3+√7)
= (3+√7)²/(9−7) = (9+6√7+7)/2 = (16+6√7)/2
= 8 + 3√7
So a = 8, b = 3.

Simplify: (2^(3/2) × 3^(−1/2)) / (6^(−3/2) × 2^(−1/2))

3 Mark

Answer:
= 2^(3/2) × 3^(−1/2) × 6^(3/2) × 2^(1/2)
= 2^(3/2+1/2) × 3^(−1/2) × (2×3)^(3/2)
= 2² × 3^(−1/2) × 2^(3/2) × 3^(3/2)
= 2^(2+3/2) × 3^(−1/2+3/2) = 2^(7/2) × 3^1 = 3 × 2^(7/2) = 3 × 8√2 = 24√2

Find two irrational numbers between √2 and √3.

3 Mark

Answer:
√2 = 1.41421... and √3 = 1.73205...
We need non-terminating non-recurring numbers in this range.
Two examples: √5/√2 = √(5/2) = √2.5 ≈ 1.581... and 1.50150015000150000...

Simplify: (√5+√3)²/(√5−√3)² leaving no surds in denominator.

3 Mark

Answer: (8+√15)/2 ... let's compute carefully:
Numerator: (√5+√3)² = 5+2√15+3 = 8+2√15
Denominator: (√5−√3)² = 5−2√15+3 = 8−2√15
= (8+2√15)/(8−2√15) × (8+2√15)/(8+2√15)
= (8+2√15)²/(64−60) = (64+32√15+60)/4 = (124+32√15)/4 = 31+8√15

If 2^x = 3^y = 6^z, prove that z = xy/(x+y).

3 Mark

Answer:
Let 2^x = 3^y = 6^z = k
Then 2 = k^(1/x), 3 = k^(1/y), 6 = k^(1/z)
Since 6 = 2×3: k^(1/z) = k^(1/x) × k^(1/y) = k^(1/x + 1/y)
1/z = 1/x + 1/y = (x+y)/(xy)
Therefore z = xy/(x+y) ✓

Show that (3√5 − 5√3)/(3√5 + 5√3) + (3√5 + 5√3)/(3√5 − 5√3) = 7⅝

3 Mark

Answer:
Let a = 3√5, b = 5√3. We need (a−b)/(a+b) + (a+b)/(a−b)
= [(a−b)² + (a+b)²] / [(a+b)(a−b)] = [2a²+2b²] / (a²−b²)
a² = 45, b² = 75
= 2(45+75)/(45−75) = 2(120)/(−30) = 240/(−30) = −8... hmm check: should be 61/8. Let's recheck the problem setup is correct — the sum of a reciprocal pair is always ≥ 2 or ≤ −2, not 7⅝. The expression evaluates to −8.

Section E — Long Answer / Proof (5 marks each)

Prove that √2 is irrational.

5 Mark

Proof by contradiction:
Assume √2 is rational. Then √2 = p/q where p, q are integers with no common factors (co-prime) and q ≠ 0.
Squaring: 2 = p²/q² → p² = 2q².
So p² is even → p is even (if p were odd, p² would be odd).
Let p = 2m. Then 4m² = 2q² → q² = 2m² → q² is even → q is even.
But p and q are both even → they share factor 2. Contradiction! (We assumed they were co-prime.)
Therefore √2 is irrational. ✓

Prove that √3 is irrational, and hence show that 2+√3 is irrational.

5 Mark

Part 1 — √3 is irrational:
Assume √3 = p/q (co-prime, q≠0). Then 3 = p²/q² → p² = 3q².
→ 3 | p² → 3 | p (prime factorisation). Let p = 3m.
Then 9m² = 3q² → q² = 3m² → 3 | q.
Both p, q divisible by 3. Contradicts co-prime assumption. ∴ √3 is irrational. ✓

Part 2 — 2+√3 is irrational:
Assume 2+√3 = r (rational). Then √3 = r−2. Since r is rational and 2 is rational, r−2 is rational.
But √3 is irrational — contradiction. Therefore 2+√3 is irrational. ✓

Find 6 rational numbers between 3/7 and 5/7. Also find 3 irrational numbers in the same range.

5 Mark

Rational numbers:
3/7 = 30/70, 5/7 = 50/70. Six rationals: 31/70, 33/70, 35/70, 37/70, 40/70, 45/70
Simplified: 31/70, 33/70, 1/2, 37/70, 4/7, 9/14

Irrational numbers:
3/7 ≈ 0.4286, 5/7 ≈ 0.7143. Need non-terminating non-recurring decimals in this range:
1. 0.4301300130001300001...
2. 0.5050050005000050000...
3. 0.6101001000100001...

Express 0.6̄, 0.47̄, 0.001̄ in the form p/q. Show all working and verify your answers.

5 Mark

0.6̄ :
x = 0.666..., 10x = 6.666... = 6+x → 9x = 6 → x = 2/3
Verify: 2/3 = 0.666... ✓

0.47̄ (= 0.4777...):
x = 0.4777..., 10x = 4.777..., 100x = 47.777...
100x − 10x = 43 → 90x = 43 → x = 43/90
Verify: 43/90 = 0.4777... ✓

0.001̄ (= 0.001001001...):
x = 0.001001..., 1000x = 1.001001... = 1+x → 999x = 1 → x = 1/999
Verify: 1/999 = 0.001001... ✓

Simplify by rationalising denominators: [1/(√5+√2)] + [1/(√5−√2)] and express in simplest form.

5 Mark

Answer: 10√5/3... let me compute:
1/(√5+√2) = (√5−√2)/(5−2) = (√5−√2)/3
1/(√5−√2) = (√5+√2)/(5−2) = (√5+√2)/3
Sum = [(√5−√2) + (√5+√2)]/3 = 2√5/3

If x = (3−√5)/2, find x² + 1/x².

5 Mark

Answer: 11
1/x = 2/(3−√5) = 2(3+√5)/((3−√5)(3+√5)) = 2(3+√5)/(9−5) = (3+√5)/2
x + 1/x = (3−√5)/2 + (3+√5)/2 = 6/2 = 3
x² + 1/x² = (x+1/x)² − 2 = 9 − 2 = 7

Show by long division that 1/17 has a repeating block of at most 16 digits. Why can't the block be longer?

5 Mark

Answer:
1/17 = 0.0588235294117647... (block of exactly 16 digits repeats)

Why ≤ 16 digits?
When dividing p by q (q = 17), the remainder at each step is one of 0, 1, 2, ..., 16 — that's only 17 possible values.
After at most 17 steps, a remainder must repeat. When it does, the decimal expansion repeats.
Since there are 16 non-zero remainders possible, the block can have at most 16 digits.
In general: For 1/q, the repeating block has at most (q−1) digits. ✓

Prove that (p+q) is irrational if p is rational and q is irrational. Use it to show 5+√3 is irrational.

5 Mark

Proof:
Let p be rational and q be irrational. Assume p+q = r (rational).
Then q = r−p. Since r is rational and p is rational, r−p is rational.
But q is irrational — this is a contradiction.
Therefore p+q is irrational. ✓

Application:
5 is rational. √3 is irrational (proved by contradiction similar to √2).
By the theorem above, 5+√3 is irrational. ✓

If a = 5+2√6 and b = 1/a, find a² + b² and a³ + b³.

5 Mark

Answer:
b = 1/(5+2√6) = (5−2√6)/((5+2√6)(5−2√6)) = (5−2√6)/(25−24) = 5−2√6
a+b = 10, ab = (5+2√6)(5−2√6) = 25−24 = 1
a² + b² = (a+b)² − 2ab = 100−2 = 98
a³ + b³ = (a+b)(a²−ab+b²) = 10(98−1) = 10×97 = 970

Prove laws of exponents: (i) aᵖ·aᵍ = a^(p+q) (ii) (aᵖ)ᵍ = a^(pq) for positive real a and rational p, q. Give two numerical examples for each.

5 Mark

(i) aᵖ·aᵍ = a^(p+q)
Write p = m/n, q = r/s (lowest terms). Then aᵖ = a^(m/n) = (a^m)^(1/n), aᵍ = (a^r)^(1/s).
aᵖ·aᵍ = a^(m/n + r/s) = a^((ms+rn)/(ns)) = a^(p+q) ✓
Examples: 2^(1/2)·2^(1/2) = 2^1 = 2 ✓ | 3^(1/3)·3^(2/3) = 3^1 = 3 ✓

(ii) (aᵖ)ᵍ = a^(pq)
((a^(m/n))^(r/s) = (a^(m/n))^(r/s) = ((ⁿ√a)^m)^(r/s) = (ⁿ√a)^(mr/s) = a^(mr/(ns)) = a^((m/n)·(r/s)) = a^(pq) ✓
Examples: (2²)³ = 2⁶ = 64 ✓ | (9^(1/2))³ = 9^(3/2) = 27 ✓