👋 Introduction
What is this chapter about? Class 10 Chapter 1 — Real Numbers — covers two powerful ideas: Euclid's Division Algorithm (for finding HCF) and the Fundamental Theorem of Arithmetic (for unique prime factorisation). You'll use these to prove numbers like √2 are irrational and understand when decimals terminate.
This chapter carries 6–8 marks in the board exam. The questions are predictable and rewarding. Master these solutions and you're guaranteed full marks on Chapter 1.
Fundamental Theorem
HCF & LCM
Irrational Numbers
Decimal Expansions
Proof by Contradiction
📐 Formula Sheet
🔢 1. Fundamental Theorem of Arithmetic
Every composite number = Unique product of primes
HCF: Product of lowest powers of common prime factors.
LCM: Product of highest powers of all prime factors.
Key: HCF(a,b) × LCM(a,b) = a × b (valid for TWO numbers only)
LCM: Product of highest powers of all prime factors.
Key: HCF(a,b) × LCM(a,b) = a × b (valid for TWO numbers only)
32760 = 2³ × 3² × 5 × 7 × 13
🔍 2. HCF × LCM Relation
HCF(a,b) × LCM(a,b) = a × b
Note: This formula works ONLY for TWO positive integers. For three numbers p, q, r — it does NOT hold directly.
Use case: If HCF is known, find LCM quickly using LCM = (a × b) ÷ HCF
Use case: If HCF is known, find LCM quickly using LCM = (a × b) ÷ HCF
HCF(6,20)=2, LCM(6,20)=60 → 2×60 = 6×20 = 120 ✓
🔑 3. Irrationality — Key Lemma
If prime p | a², then p | a
Method: Proof by Contradiction — assume the number is rational (= a/b in lowest terms), then derive that a prime divides both a and b, contradicting HCF(a,b) = 1.
Rule: Sum/difference of rational + irrational = irrational
Rule: Sum/difference of rational + irrational = irrational
√2, √3, √5, 5−√3, 3√2 — all irrational
📊 4. Decimal Expansions
p/q terminates ↔ q = 2ⁿ × 5ᵐ
Terminating: Denominator (in lowest terms) has only factors 2 and/or 5.
Non-terminating repeating: Denominator has any other prime factor (3, 7, 11…)
Irrational: Non-terminating, non-repeating decimal
Non-terminating repeating: Denominator has any other prime factor (3, 7, 11…)
Irrational: Non-terminating, non-repeating decimal
7/8 = 7/2³ → terminates (0.875) | 1/6 → non-terminating
📖 NCERT Solutions — All Examples
📘 Section 1.2 — Fundamental Theorem Examples
Example 1: For 4ⁿ (n = natural number) — does it ever end with digit 0?
Conceptual+
Question: Check whether 4ⁿ ends with digit 0 for any natural number n.
Solution:
For any number to end with 0, it must be divisible by 10.
Since 10 = 2 × 5, the prime factorisation must contain the prime 5.
Now, 4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime factor of 4ⁿ is 2. It does NOT contain 5.
By the uniqueness of the Fundamental Theorem of Arithmetic, there is no way to get 5 as a factor of 4ⁿ.
For any number to end with 0, it must be divisible by 10.
Since 10 = 2 × 5, the prime factorisation must contain the prime 5.
Now, 4ⁿ = (2²)ⁿ = 2²ⁿ
The only prime factor of 4ⁿ is 2. It does NOT contain 5.
By the uniqueness of the Fundamental Theorem of Arithmetic, there is no way to get 5 as a factor of 4ⁿ.
∴ There is NO natural number n for which 4ⁿ ends with digit 0.
Example 2: Find LCM and HCF of 6 and 20 by prime factorisation.
2 marks+
Find LCM and HCF of 6 and 20 by prime factorisation method.
Step 1 — Prime Factorise:
6 = 2¹ × 3¹
20 = 2² × 5¹
Step 2 — HCF = Product of LOWEST powers of COMMON prime factors
Common prime factor = 2 → lowest power = 2¹
HCF(6, 20) = 2¹ = 2
Step 3 — LCM = Product of HIGHEST powers of ALL prime factors
Primes: 2, 3, 5 → highest powers: 2², 3¹, 5¹
LCM(6, 20) = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
Verification: HCF × LCM = 2 × 60 = 120 = 6 × 20 ✓
6 = 2¹ × 3¹
20 = 2² × 5¹
Step 2 — HCF = Product of LOWEST powers of COMMON prime factors
Common prime factor = 2 → lowest power = 2¹
HCF(6, 20) = 2¹ = 2
Step 3 — LCM = Product of HIGHEST powers of ALL prime factors
Primes: 2, 3, 5 → highest powers: 2², 3¹, 5¹
LCM(6, 20) = 2² × 3¹ × 5¹ = 4 × 3 × 5 = 60
Verification: HCF × LCM = 2 × 60 = 120 = 6 × 20 ✓
HCF(6, 20) = 2 | LCM(6, 20) = 60
Example 3: HCF of 96 and 404 → find LCM.
3 marks+
Find HCF of 96 and 404 by prime factorisation. Hence find LCM.
Prime Factorise:
96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3
96 = 2⁵ × 3¹
404 = 2 × 202 = 2 × 2 × 101
404 = 2² × 101 (101 is prime)
HCF = product of lowest powers of common factors
Common factor: 2 → lowest power = 2²
HCF(96, 404) = 2² = 4
LCM using formula: LCM = (a × b) ÷ HCF
LCM = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696
96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3
96 = 2⁵ × 3¹
404 = 2 × 202 = 2 × 2 × 101
404 = 2² × 101 (101 is prime)
HCF = product of lowest powers of common factors
Common factor: 2 → lowest power = 2²
HCF(96, 404) = 2² = 4
LCM using formula: LCM = (a × b) ÷ HCF
LCM = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696
HCF(96, 404) = 4 | LCM(96, 404) = 9696
Example 4: HCF and LCM of 6, 72 and 120.
3 marks+
Find HCF and LCM of 6, 72 and 120 using prime factorisation.
Prime Factorise all three:
6 = 2¹ × 3¹
72 = 2³ × 3²
120 = 2³ × 3¹ × 5¹
HCF = lowest powers of COMMON primes (2 and 3 are common to all)
Common primes: 2 → min power = 2¹; 3 → min power = 3¹
HCF(6, 72, 120) = 2¹ × 3¹ = 6
LCM = highest powers of ALL primes (2, 3, 5)
2 → max power = 2³; 3 → max power = 3²; 5 → max power = 5¹
LCM(6, 72, 120) = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360
Remark: 6 × 72 × 120 = 51,840 ≠ HCF × LCM = 6 × 360 = 2160
∴ HCF × LCM ≠ product of three numbers (formula only for two numbers!)
6 = 2¹ × 3¹
72 = 2³ × 3²
120 = 2³ × 3¹ × 5¹
HCF = lowest powers of COMMON primes (2 and 3 are common to all)
Common primes: 2 → min power = 2¹; 3 → min power = 3¹
HCF(6, 72, 120) = 2¹ × 3¹ = 6
LCM = highest powers of ALL primes (2, 3, 5)
2 → max power = 2³; 3 → max power = 3²; 5 → max power = 5¹
LCM(6, 72, 120) = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360
Remark: 6 × 72 × 120 = 51,840 ≠ HCF × LCM = 6 × 360 = 2160
∴ HCF × LCM ≠ product of three numbers (formula only for two numbers!)
HCF(6, 72, 120) = 6 | LCM(6, 72, 120) = 360
❓ Stuck on a question?
Ask your doubt here 👉
📘 Section 1.3 — Irrational Numbers Examples
Example 5 (Theorem 1.3): Prove that √2 is irrational.
3 marks ★★★+
Prove that √2 is irrational. (Proof by Contradiction)
Assume: Let √2 be rational.
Then √2 = a/b, where a, b are integers, b ≠ 0, and HCF(a, b) = 1 (i.e., a/b is in lowest terms).
Step 1: √2 = a/b → Squaring both sides → 2 = a²/b² → a² = 2b²
Step 2: Since a² = 2b², we know 2 divides a².
By Theorem 1.2 (key lemma: if p | a², then p | a), since 2 is prime:
→ 2 divides a
Step 3: Since 2 | a, write a = 2c for some integer c.
Substitute: a² = 2b² → (2c)² = 2b² → 4c² = 2b² → b² = 2c²
Step 4: Since b² = 2c², we know 2 divides b². By the same lemma:
→ 2 divides b
Contradiction: Both a and b are divisible by 2.
But we assumed HCF(a, b) = 1. This is a contradiction!
Then √2 = a/b, where a, b are integers, b ≠ 0, and HCF(a, b) = 1 (i.e., a/b is in lowest terms).
Step 1: √2 = a/b → Squaring both sides → 2 = a²/b² → a² = 2b²
Step 2: Since a² = 2b², we know 2 divides a².
By Theorem 1.2 (key lemma: if p | a², then p | a), since 2 is prime:
→ 2 divides a
Step 3: Since 2 | a, write a = 2c for some integer c.
Substitute: a² = 2b² → (2c)² = 2b² → 4c² = 2b² → b² = 2c²
Step 4: Since b² = 2c², we know 2 divides b². By the same lemma:
→ 2 divides b
Contradiction: Both a and b are divisible by 2.
But we assumed HCF(a, b) = 1. This is a contradiction!
∴ Our assumption was wrong. √2 is irrational. □
Example 5 (Textbook): Prove that √3 is irrational.
3 marks+
Prove that √3 is irrational.
Assume: Let √3 be rational → √3 = a/b (in lowest terms, HCF(a,b) = 1).
Step 1: b√3 = a → Squaring → 3b² = a²
→ 3 divides a² → By lemma (3 is prime) → 3 divides a
Step 2: Write a = 3c. Then 3b² = (3c)² = 9c² → b² = 3c²
→ 3 divides b² → 3 divides b
Contradiction: Both a and b divisible by 3, contradicts HCF(a,b) = 1.
Step 1: b√3 = a → Squaring → 3b² = a²
→ 3 divides a² → By lemma (3 is prime) → 3 divides a
Step 2: Write a = 3c. Then 3b² = (3c)² = 9c² → b² = 3c²
→ 3 divides b² → 3 divides b
Contradiction: Both a and b divisible by 3, contradicts HCF(a,b) = 1.
∴ √3 is irrational. □
Example 6: Show that 5 − √3 is irrational.
3 marks+
Show that 5 − √3 is irrational.
Assume: Let 5 − √3 be rational.
Then 5 − √3 = a/b (where a, b integers, b ≠ 0, HCF(a,b) = 1)
Rearrange: √3 = 5 − a/b = (5b − a)/b
Since a and b are integers, (5b − a)/b is a rational number.
So √3 would be rational.
Contradiction: But √3 is irrational (proven above). This is a contradiction.
Then 5 − √3 = a/b (where a, b integers, b ≠ 0, HCF(a,b) = 1)
Rearrange: √3 = 5 − a/b = (5b − a)/b
Since a and b are integers, (5b − a)/b is a rational number.
So √3 would be rational.
Contradiction: But √3 is irrational (proven above). This is a contradiction.
∴ 5 − √3 is irrational. □
Example 7: Show that 3√2 is irrational.
3 marks+
Show that 3√2 is irrational.
Assume: Let 3√2 be rational → 3√2 = a/b (in lowest terms)
Rearrange: √2 = a/(3b)
Since a, b, 3 are integers, a/(3b) is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
Rearrange: √2 = a/(3b)
Since a, b, 3 are integers, a/(3b) is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
∴ 3√2 is irrational. □
📝 Exercise 1.1 — All Solutions
📗 Exercise 1.1 — Prime Factorisation, HCF & LCM
Q1. Express each number as a product of prime factors: 140, 156, 3825, 5005, 7429
1 mark each+
(i) 140
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7
140 = 2² × 5 × 7
(ii) 156
156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13
156 = 2² × 3 × 13
(iii) 3825
3825 ÷ 3 = 1275; 1275 ÷ 3 = 425; 425 ÷ 5 = 85; 85 ÷ 5 = 17
3825 = 3² × 5² × 17
(iv) 5005
5005 ÷ 5 = 1001; 1001 ÷ 7 = 143; 143 ÷ 11 = 13
5005 = 5 × 7 × 11 × 13
(v) 7429
7429 ÷ 17 = 437; 437 ÷ 19 = 23
7429 = 17 × 19 × 23
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7
140 = 2² × 5 × 7
(ii) 156
156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13
156 = 2² × 3 × 13
(iii) 3825
3825 ÷ 3 = 1275; 1275 ÷ 3 = 425; 425 ÷ 5 = 85; 85 ÷ 5 = 17
3825 = 3² × 5² × 17
(iv) 5005
5005 ÷ 5 = 1001; 1001 ÷ 7 = 143; 143 ÷ 11 = 13
5005 = 5 × 7 × 11 × 13
(v) 7429
7429 ÷ 17 = 437; 437 ÷ 19 = 23
7429 = 17 × 19 × 23
Q2. Find LCM and HCF and verify LCM × HCF = product: (i) 26 & 91 (ii) 510 & 92 (iii) 336 & 54
3 marks each+
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13 (common: 13¹)
LCM = 2 × 7 × 13 = 182
Verify: 13 × 182 = 2366 = 26 × 91 = 2366 ✓
(ii) 510 and 92
510 = 2 × 255 = 2 × 3 × 85 = 2 × 3 × 5 × 17
92 = 2 × 46 = 2 × 2 × 23 = 2² × 23
HCF = 2¹ = 2 (common: 2, lowest power)
LCM = 2² × 3 × 5 × 17 × 23 = 4 × 3 × 5 × 17 × 23 = 23460
Verify: 2 × 23460 = 46920 = 510 × 92 = 46920 ✓
(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 27 = 2 × 3³
HCF = 2¹ × 3¹ = 6 (common: 2 and 3, lowest powers)
LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024
Verify: 6 × 3024 = 18144 = 336 × 54 = 18144 ✓
26 = 2 × 13
91 = 7 × 13
HCF = 13 (common: 13¹)
LCM = 2 × 7 × 13 = 182
Verify: 13 × 182 = 2366 = 26 × 91 = 2366 ✓
(ii) 510 and 92
510 = 2 × 255 = 2 × 3 × 85 = 2 × 3 × 5 × 17
92 = 2 × 46 = 2 × 2 × 23 = 2² × 23
HCF = 2¹ = 2 (common: 2, lowest power)
LCM = 2² × 3 × 5 × 17 × 23 = 4 × 3 × 5 × 17 × 23 = 23460
Verify: 2 × 23460 = 46920 = 510 × 92 = 46920 ✓
(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 27 = 2 × 3³
HCF = 2¹ × 3¹ = 6 (common: 2 and 3, lowest powers)
LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024
Verify: 6 × 3024 = 18144 = 336 × 54 = 18144 ✓
Q3. Find LCM and HCF by prime factorisation: (i) 12,15,21 (ii) 17,23,29 (iii) 8,9,25
3 marks each+
(i) 12, 15 and 21
12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7
HCF = 3¹ = 3 (only 3 is common to all, min power = 1)
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
17, 23, 29 are all prime numbers.
HCF = 1 (no common factor)
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
8 = 2³; 9 = 3²; 25 = 5²
No common prime factor → HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7
HCF = 3¹ = 3 (only 3 is common to all, min power = 1)
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
17, 23, 29 are all prime numbers.
HCF = 1 (no common factor)
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
8 = 2³; 9 = 3²; 25 = 5²
No common prime factor → HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
Q4. Given HCF(306, 657) = 9, find LCM(306, 657).
2 marks+
Using: HCF × LCM = a × b
9 × LCM = 306 × 657
9 × LCM = 201042
LCM = 201042 ÷ 9 = 22338
9 × LCM = 306 × 657
9 × LCM = 201042
LCM = 201042 ÷ 9 = 22338
LCM(306, 657) = 22338
Q5. Check whether 6ⁿ can end with digit 0 for any natural number n.
2 marks+
For 6ⁿ to end in 0, it must be divisible by 10 = 2 × 5.
So prime factorisation of 6ⁿ must contain the prime 5.
Now: 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factors of 6ⁿ are only 2 and 3. There is no factor of 5.
By uniqueness of the Fundamental Theorem, 5 can never be a factor of 6ⁿ.
So prime factorisation of 6ⁿ must contain the prime 5.
Now: 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factors of 6ⁿ are only 2 and 3. There is no factor of 5.
By uniqueness of the Fundamental Theorem, 5 can never be a factor of 6ⁿ.
∴ 6ⁿ can NEVER end with digit 0 for any natural number n.
Q6. Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.
2 marks+
Part (a): 7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 2 × 39 = 13 × 2 × 3 × 13
This has factors other than 1 and itself → Composite ✓
Part (b): 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
This has factor 5 (and 1009) → Composite ✓
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 2 × 39 = 13 × 2 × 3 × 13
This has factors other than 1 and itself → Composite ✓
Part (b): 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
This has factor 5 (and 1009) → Composite ✓
Both numbers are composite because they have factors other than 1 and themselves.
Q7. Sonia takes 18 min, Ravi takes 12 min to complete one round. After how many minutes will they meet at starting point?
2 marks+
They will meet again at the starting point after LCM(18, 12) minutes.
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36 minutes
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36 minutes
They will meet again at the starting point after 36 minutes.
📝 Exercise 1.2 — All Solutions
📗 Exercise 1.2 — Irrational Numbers
Q1. Prove that √5 is irrational.
3 marks ★★★+
Prove that √5 is irrational. (Most important exam question!)
Assume: Let √5 be rational → √5 = a/b (in lowest terms, so HCF(a,b) = 1)
Step 1: Squaring → 5 = a²/b² → a² = 5b²
Since 5 divides a², and 5 is prime → by Theorem 1.2: 5 divides a
Step 2: Write a = 5k for some integer k.
Substitute: a² = 5b² → (5k)² = 5b² → 25k² = 5b² → b² = 5k²
→ 5 divides b² → 5 divides b
Contradiction: 5 divides both a and b. But HCF(a,b) = 1. Contradiction!
Step 1: Squaring → 5 = a²/b² → a² = 5b²
Since 5 divides a², and 5 is prime → by Theorem 1.2: 5 divides a
Step 2: Write a = 5k for some integer k.
Substitute: a² = 5b² → (5k)² = 5b² → 25k² = 5b² → b² = 5k²
→ 5 divides b² → 5 divides b
Contradiction: 5 divides both a and b. But HCF(a,b) = 1. Contradiction!
∴ Our assumption is wrong. √5 is irrational. □
Q2. Prove that 3 + 2√5 is irrational.
3 marks ★★★+
Prove that 3 + 2√5 is irrational. (Very frequently asked!)
Assume: Let 3 + 2√5 be rational.
Then 3 + 2√5 = a/b (where a, b are integers, b ≠ 0, HCF(a,b) = 1)
Rearrange:
2√5 = a/b − 3
2√5 = (a − 3b)/b
√5 = (a − 3b) / (2b)
Since a, b are integers, (a − 3b)/(2b) is a rational number.
So √5 would be rational.
Contradiction: But √5 is irrational (proved in Q1). Contradiction!
Then 3 + 2√5 = a/b (where a, b are integers, b ≠ 0, HCF(a,b) = 1)
Rearrange:
2√5 = a/b − 3
2√5 = (a − 3b)/b
√5 = (a − 3b) / (2b)
Since a, b are integers, (a − 3b)/(2b) is a rational number.
So √5 would be rational.
Contradiction: But √5 is irrational (proved in Q1). Contradiction!
∴ 3 + 2√5 is irrational. □
Q3(i). Prove that 1/√2 is irrational.
3 marks+
Assume: Let 1/√2 be rational → 1/√2 = a/b (HCF(a,b) = 1)
Rearrange: √2 = b/a
Since a, b are integers and a ≠ 0, b/a is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
Rearrange: √2 = b/a
Since a, b are integers and a ≠ 0, b/a is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
∴ 1/√2 is irrational. □
Q3(ii). Prove that 7√5 is irrational.
3 marks+
Assume: Let 7√5 be rational → 7√5 = a/b (HCF(a,b) = 1)
Rearrange: √5 = a/(7b)
Since a, b, 7 are integers, a/(7b) is rational → √5 is rational.
Contradiction: √5 is irrational. Contradiction!
Rearrange: √5 = a/(7b)
Since a, b, 7 are integers, a/(7b) is rational → √5 is rational.
Contradiction: √5 is irrational. Contradiction!
∴ 7√5 is irrational. □
Q3(iii). Prove that 6 + √2 is irrational.
3 marks+
Assume: Let 6 + √2 be rational → 6 + √2 = a/b (HCF(a,b) = 1)
Rearrange: √2 = a/b − 6 = (a − 6b)/b
Since a, b are integers, (a − 6b)/b is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
Rearrange: √2 = a/b − 6 = (a − 6b)/b
Since a, b are integers, (a − 6b)/b is rational → √2 is rational.
Contradiction: √2 is irrational. Contradiction!
∴ 6 + √2 is irrational. □
📊 Quick Reference Table
| Concept | Formula / Rule | Example |
|---|---|---|
| Prime Factorisation | Every composite = unique product of primes | 360 = 2³×3²×5 |
| HCF (2 numbers) | Lowest powers of common primes | HCF(12,18)=6 |
| LCM (2 numbers) | Highest powers of all primes | LCM(12,18)=36 |
| HCF × LCM | = a × b (TWO numbers only!) | 6×36=12×18 ✓ |
| Irrationality proof | Contradiction: assume rational, derive paradox | √2, √3, √5, 3+2√5 |
| Key Lemma | p prime, p|a² → p|a | 2|a² → 2|a |
| Terminating decimal | q = 2ⁿ × 5ᵐ | 7/8→0.875 ✓ |
| Non-terminating | q has prime other than 2 or 5 | 1/3=0.333… |
| 4ⁿ ends in 0? | Never — 4ⁿ=2²ⁿ has no factor 5 | No natural n |
| 6ⁿ ends in 0? | Never — 6ⁿ=2ⁿ×3ⁿ has no factor 5 | No natural n |
⚠️ Common Mistakes
❌ Wrong
HCF × LCM = a×b×c for three numbers
✅ Correct
HCF × LCM = a×b ONLY for TWO numbers
❌ Wrong
LCM = product of lowest powers
✅ Correct
LCM = HIGHEST powers; HCF = LOWEST powers
❌ Wrong
Not stating "HCF(a,b) = 1" at start of proof
✅ Correct
Always begin: "Let a/b be in lowest terms, i.e., HCF(a,b) = 1"
❌ Wrong
Checking 13/91 termination: denominator 91 = 7×13 → non-terminating!?
✅ Correct
First reduce: 13/91 = 1/7. Denominator 7 ≠ 2ⁿ×5ᵐ → non-terminating
🎯 Exam Tips from Manish Sir
- Irrationality proofs = guaranteed 3 marksCBSE asks √5, 3+2√5, 7√5 type every year. Practice all 5 variants until you can write blindfolded.
- Always write every stepFor 3-mark proofs, each logical step = marks. Never skip "let a = 2c" or "substituting".
- Always verify HCF × LCM = a × bExaminers love verification. Add one line at the end — it can save you from losing a mark.
- Chapter 1 = easiest 7 marksQuestions here repeat every year. Do all NCERT examples + both exercises — you're done!
- Reduce fraction before decimal checkAlways simplify p/q to lowest terms before checking denominator for termination.
❓ Frequently Asked Questions
What is the Fundamental Theorem of Arithmetic?
Every composite number can be expressed as a product of primes in a unique way (ignoring order). E.g., 360 = 2³ × 3² × 5. This uniqueness is what makes HCF/LCM by prime factorisation work.
Can HCF × LCM = a × b × c for three numbers?
No! This formula is valid only for two positive integers. For three numbers p, q, r, the formula becomes more complex and different. This is a very common mistake in exams.
How many marks does Real Numbers carry in Class 10 board?
Chapter 1 typically carries 6 to 8 marks — usually 1 MCQ (1 mark) + 1 short answer (2 marks) + 1 long answer (3 marks). All questions are highly predictable.
What is the difference between HCF and LCM in prime factorisation?
HCF = product of lowest (minimum) powers of common prime factors only. LCM = product of highest (maximum) powers of all prime factors (common + non-common).
What is the key lemma used in all irrational proofs?
Theorem 1.2: If p is a prime number and p divides a², then p divides a. This single statement enables all irrationality proofs. Memorise it!
Is π rational or irrational?
π is irrational — non-terminating, non-repeating (3.14159…). Note: 22/7 ≈ 3.142857 is only an approximation. 22/7 ≠ π.
🔗 Explore More
🎯 Want Complete Maths Preparation?
Join our WhatsApp group and get:
- Chapter-wise Notes
- PYQs (Last 10 Years)
- Important Questions
- Practice Papers
- Doubt Solving Support
🔥 Limited access for Board 2027 students — Join with Manish Sir's approval
👉 Join Now: wa.me/918107563723📌 You will be added by Manish Sir after a quick approval