Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Text lesson
Chapter 1. Real Numbers
Grade 10 Mathematics · CBSE / NCERT · Chapter 1
Real Numbers
Master the Fundamental Theorem of Arithmetic, prime factorisation, HCF & LCM, and the elegant proof-by-contradiction that makes irrational numbers forever irrational.
| 🔢 Prime Factorisation | 📐 HCF & LCM | ∞ Irrational Numbers | 📝 Proof by Contradiction |
| 📋 Chapter at a Glance |
| Section | Topics Covered | Key Results |
| 1.2 Fundamental Theorem of Arithmetic | Prime factorisation of composite numbers, factor tree method, uniqueness of factorisation, applications to HCF and LCM | Every composite number = unique product of primes; HCF × LCM = a × b |
| 1.3 Irrational Numbers | Proof that √2, √3, √5 are irrational; proof by contradiction method; showing combinations like 5−√3 and 3√2 are irrational | If p is prime and p | a², then p | a; √p is irrational for any prime p |
| 🔑 Key Definitions |
Prime Number: A natural number greater than 1 that has no positive divisors other than 1 and itself. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23… Note: 2 is the only even prime number.
Composite Number: A natural number greater than 1 that has at least one positive divisor other than 1 and itself. In other words, it can be expressed as a product of two or more prime numbers. Examples: 4, 6, 8, 9, 10, 12…
Prime Factorisation: Expressing a composite number as a product of its prime factors (written in ascending order of the prime factors). This representation is unique — there is one and only one way to write a number as a product of primes (apart from order). Example: 360 = 2³ × 3² × 5.
HCF (Highest Common Factor): The largest number that divides two or more given numbers exactly (without remainder). Also called GCD (Greatest Common Divisor). HCF = product of the smallest power of each common prime factor in the numbers.
LCM (Lowest Common Multiple): The smallest positive integer that is divisible by two or more given numbers. LCM = product of the greatest power of each prime factor involved in any of the numbers. For two numbers a, b: HCF(a,b) × LCM(a,b) = a × b.
Rational Number: A number that can be written in the form p/q, where p and q are integers and q ≠ 0. The decimal expansion of a rational number is either terminating or non-terminating repeating. Examples: 3/4, −7/2, 0.5, 0.333…
Irrational Number: A number that 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-repeating. Examples: √2, √3, √5, π, 0.10110111011110…
Coprime Numbers: Two integers are said to be coprime (or relatively prime) if their HCF is 1 — they share no common factor other than 1. Example: 8 and 15 are coprime (HCF = 1). Note: coprime numbers need not be prime themselves.
Proof by Contradiction: A mathematical proof technique where we assume the opposite (negation) of what we want to prove, then show that this assumption leads to a logical contradiction. Since a contradiction cannot be true, our original assumption must be false, so what we wanted to prove must be true.
Factor Tree: A visual diagram that breaks down a composite number into its prime factors by repeated division. Starting from the number, we branch out by dividing by the smallest possible prime factor at each step until all branches end in prime numbers.
| 1.2 — The Fundamental Theorem of Arithmetic |
Every natural number can be written as a product of its prime factors. The remarkable fact — known as the Fundamental Theorem of Arithmetic — is that this factorisation is always unique. No matter how you factorise a number, you always get the same set of prime factors (just possibly in a different order).
Theorem 1.1 — Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
| 📐 Diagram 1: Factor Tree for 32760 |
32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13
No matter how you build the factor tree, you always get the same prime factors — this is the uniqueness guaranteed by the Fundamental Theorem. |
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| HCF and LCM by Prime Factorisation Method |
Once we have the prime factorisations of two or more numbers, finding HCF and LCM becomes straightforward using these rules:
| What to find | Rule | Example: 6 = 2¹×3¹, 20 = 2²×5¹ |
| HCF | Product of smallest power of each common prime factor | Common prime: 2. Smallest power: 2¹. HCF = 2 |
| LCM | Product of greatest power of each prime factor (from any number) | All primes: 2,3,5. Greatest powers: 2²,3¹,5¹. LCM = 4×3×5 = 60 |
| Key Relation | HCF(a, b) × LCM(a, b) = a × b [Valid for TWO numbers only. For three numbers: HCF × LCM ≠ a × b × c] | |
| 📐 Diagram 2: HCF and LCM — Visual Comparison (6, 72, 120) |
Note: For three numbers, HCF × LCM ≠ 6 × 72 × 120. The HCF × LCM = a × b rule only holds for exactly two numbers. |
| 1.3 — Revisiting Irrational Numbers |
An irrational number cannot be written as p/q. We prove irrationality using proof by contradiction: assume the number IS rational, then show that leads to an impossible contradiction.
Theorem 1.2 (Key Lemma): Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
This theorem is the engine behind all the irrational number proofs.
This theorem is the engine behind all the irrational number proofs.
| 📐 Diagram 3: The Proof by Contradiction Framework |
|
| Rules for Rational + Irrational Combinations |
| Operation | Rule | Example |
| Rational ± Irrational | Always IRRATIONAL | 5 − √3 is irrational; 2 + √5 is irrational |
| Non-zero Rational × Irrational | Always IRRATIONAL | 3√2 is irrational; (2/5)√7 is irrational |
| Rational ÷ Irrational (or vice versa) | Always IRRATIONAL | 1/√2 is irrational; √5/3 is irrational |
| 0 × Irrational | = 0 (RATIONAL) | 0 × √2 = 0 — the exception to the rule above! |
| ✏ Worked Examples |
EXAMPLE 1
Express 140 as a product of its prime factors.
Show Solution ▶
140 ÷ 2 = 70 → 70 ÷ 2 = 35 → 35 ÷ 5 = 7 → 7 is prime.
140 = 2² × 5 × 7
Check: 4 × 5 × 7 = 4 × 35 = 140 ✓
140 = 2² × 5 × 7
Check: 4 × 5 × 7 = 4 × 35 = 140 ✓
EXAMPLE 2
Find the LCM and HCF of 26 and 91 by prime factorisation. Verify that LCM × HCF = product of the two numbers.
Show Solution ▶
26 = 2 × 13; 91 = 7 × 13
Common prime factor: 13. Smallest power of 13: 13¹.
HCF(26, 91) = 13
All primes: 2, 7, 13. Greatest powers: 2¹, 7¹, 13¹.
LCM(26, 91) = 2 × 7 × 13 = 182
Verification: HCF × LCM = 13 × 182 = 2366; Product = 26 × 91 = 2366 ✓
Common prime factor: 13. Smallest power of 13: 13¹.
HCF(26, 91) = 13
All primes: 2, 7, 13. Greatest powers: 2¹, 7¹, 13¹.
LCM(26, 91) = 2 × 7 × 13 = 182
Verification: HCF × LCM = 13 × 182 = 2366; Product = 26 × 91 = 2366 ✓
EXAMPLE 3
Find HCF of 96 and 404 by prime factorisation. Hence find their LCM.
Show Solution ▶
96 = 2⁵ × 3; 404 = 2² × 101
Common prime factor: 2. Smallest power: 2².
HCF(96, 404) = 2² = 4
LCM = (96 × 404) ÷ HCF = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696
Check: 4 × 9696 = 38784 = 96 × 404 ✓
Common prime factor: 2. Smallest power: 2².
HCF(96, 404) = 2² = 4
LCM = (96 × 404) ÷ HCF = (96 × 404) ÷ 4 = 38784 ÷ 4 = 9696
Check: 4 × 9696 = 38784 = 96 × 404 ✓
EXAMPLE 4
Find HCF and LCM of 6, 72 and 120.
Show Solution ▶
6 = 2¹ × 3¹; 72 = 2³ × 3²; 120 = 2³ × 3¹ × 5¹
Common primes to ALL three: 2 and 3. Smallest powers: 2¹, 3¹.
HCF(6, 72, 120) = 2¹ × 3¹ = 6
Greatest powers of all primes (2, 3, 5): 2³, 3², 5¹.
LCM(6, 72, 120) = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360
Note: 6 × 72 × 120 = 51840 ≠ 6 × 360 = 2160. The HCF × LCM = product rule does NOT apply for three numbers.
Common primes to ALL three: 2 and 3. Smallest powers: 2¹, 3¹.
HCF(6, 72, 120) = 2¹ × 3¹ = 6
Greatest powers of all primes (2, 3, 5): 2³, 3², 5¹.
LCM(6, 72, 120) = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360
Note: 6 × 72 × 120 = 51840 ≠ 6 × 360 = 2160. The HCF × LCM = product rule does NOT apply for three numbers.
EXAMPLE 5
Check whether 4ⁿ can end with the digit 0 for any natural number n.
Show Solution ▶
A number ends in 0 if and only if it is divisible by 10 = 2 × 5, i.e., its prime factorisation must contain the prime 5.
4ⁿ = (2²)ⁿ = 2²ⁿ — the prime factorisation of 4ⁿ contains only the prime 2.
By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other primes in the factorisation of 4ⁿ. So the prime 5 never appears.
Therefore, 4ⁿ can never end with the digit 0 for any natural number n.
4ⁿ = (2²)ⁿ = 2²ⁿ — the prime factorisation of 4ⁿ contains only the prime 2.
By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other primes in the factorisation of 4ⁿ. So the prime 5 never appears.
Therefore, 4ⁿ can never end with the digit 0 for any natural number n.
EXAMPLE 6
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Show Solution ▶
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ — the prime factorisation of 6ⁿ contains only the primes 2 and 3.
For a number to end in 0, its prime factorisation must contain the prime 5 (since 10 = 2 × 5).
Since 5 never appears in the factorisation of 6ⁿ (by uniqueness of prime factorisation), there is no n for which 6ⁿ ends with digit 0.
6ⁿ can never end with the digit 0.
Note: This is different from 6ⁿ ending in 6 — that’s true for all n ≥ 1.
For a number to end in 0, its prime factorisation must contain the prime 5 (since 10 = 2 × 5).
Since 5 never appears in the factorisation of 6ⁿ (by uniqueness of prime factorisation), there is no n for which 6ⁿ ends with digit 0.
6ⁿ can never end with the digit 0.
Note: This is different from 6ⁿ ending in 6 — that’s true for all n ≥ 1.
EXAMPLE 7
Given that HCF(306, 657) = 9, find LCM(306, 657).
Show Solution ▶
Using the relation: HCF(a, b) × LCM(a, b) = a × b
9 × LCM(306, 657) = 306 × 657
9 × LCM = 201042
LCM(306, 657) = 201042 ÷ 9 = 22338
9 × LCM(306, 657) = 306 × 657
9 × LCM = 201042
LCM(306, 657) = 201042 ÷ 9 = 22338
EXAMPLE 8
Prove that √3 is irrational.
Show Solution ▶
Assume, to the contrary, that √3 is rational.
Then we can find coprime integers a and b (b ≠ 0) such that √3 = a/b.
So b√3 = a. Squaring: 3b² = a².
Therefore 3 divides a². By Theorem 1.2 (with p = 3), 3 divides a.
So a = 3c for some integer c.
Substituting: 3b² = 9c² → b² = 3c².
So 3 divides b². By Theorem 1.2, 3 divides b.
But now 3 divides both a and b — contradicting that a and b are coprime.
Conclusion: Our assumption was wrong. Therefore √3 is irrational.
Then we can find coprime integers a and b (b ≠ 0) such that √3 = a/b.
So b√3 = a. Squaring: 3b² = a².
Therefore 3 divides a². By Theorem 1.2 (with p = 3), 3 divides a.
So a = 3c for some integer c.
Substituting: 3b² = 9c² → b² = 3c².
So 3 divides b². By Theorem 1.2, 3 divides b.
But now 3 divides both a and b — contradicting that a and b are coprime.
Conclusion: Our assumption was wrong. Therefore √3 is irrational.
EXAMPLE 9
Prove that √5 is irrational.
Show Solution ▶
Assume √5 is rational. Then √5 = a/b where a, b are coprime integers, b ≠ 0.
Squaring: 5 = a²/b² → 5b² = a².
So 5 divides a². By Theorem 1.2 (p = 5), 5 divides a. Write a = 5c.
Substituting: 5b² = 25c² → b² = 5c².
So 5 divides b² → 5 divides b.
Both a and b divisible by 5 — contradicts coprime assumption.
Therefore √5 is irrational.
Squaring: 5 = a²/b² → 5b² = a².
So 5 divides a². By Theorem 1.2 (p = 5), 5 divides a. Write a = 5c.
Substituting: 5b² = 25c² → b² = 5c².
So 5 divides b² → 5 divides b.
Both a and b divisible by 5 — contradicts coprime assumption.
Therefore √5 is irrational.
EXAMPLE 10
Show that 5 − √3 is irrational.
Show Solution ▶
Assume 5 − √3 is rational. Then 5 − √3 = a/b for coprime integers a, b (b ≠ 0).
Rearranging: √3 = 5 − a/b = (5b − a)/b.
Since a and b are integers, (5b − a)/b is rational — so √3 would be rational.
But this contradicts the fact that √3 is irrational.
Therefore our assumption was wrong. 5 − √3 is irrational.
Rearranging: √3 = 5 − a/b = (5b − a)/b.
Since a and b are integers, (5b − a)/b is rational — so √3 would be rational.
But this contradicts the fact that √3 is irrational.
Therefore our assumption was wrong. 5 − √3 is irrational.
EXAMPLE 11
Show that 3√2 is irrational.
Show Solution ▶
Assume 3√2 is rational. Then 3√2 = a/b for coprime integers a, b (b ≠ 0).
Rearranging: √2 = a/(3b).
Since 3, a and b are integers, a/(3b) is rational — so √2 would be rational.
But this contradicts the fact that √2 is irrational.
Therefore our assumption was wrong. 3√2 is irrational.
Rearranging: √2 = a/(3b).
Since 3, a and b are integers, a/(3b) is rational — so √2 would be rational.
But this contradicts the fact that √2 is irrational.
Therefore our assumption was wrong. 3√2 is irrational.
EXAMPLE 12
Explain why 7 × 11 × 13 + 13 is a composite number.
Show Solution ▶
7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1) = 13 × (77 + 1) = 13 × 78
Since the number = 13 × 78, it has 13 as a factor other than 1 and itself.
Therefore it is a composite number.
(A number with a factor other than 1 and itself cannot be prime — it is composite.)
Since the number = 13 × 78, it has 13 as a factor other than 1 and itself.
Therefore it is a composite number.
(A number with a factor other than 1 and itself cannot be prime — it is composite.)
EXAMPLE 13
Sonia takes 18 min and Ravi takes 12 min to complete one round of a circular track. They start at the same time and same point. After how many minutes will they meet again at the starting point?
Show Solution ▶
They will meet again at the starting point after a time that is a common multiple of both 18 and 12. The first time they meet is at the LCM.
18 = 2 × 3²; 12 = 2² × 3
LCM(18, 12) = 2² × 3² = 4 × 9 = 36 minutes
At 36 minutes: Sonia completes 36/18 = 2 rounds; Ravi completes 36/12 = 3 rounds. Both are back at the starting point simultaneously.
18 = 2 × 3²; 12 = 2² × 3
LCM(18, 12) = 2² × 3² = 4 × 9 = 36 minutes
At 36 minutes: Sonia completes 36/18 = 2 rounds; Ravi completes 36/12 = 3 rounds. Both are back at the starting point simultaneously.
EXAMPLE 14
Prove that 3 + 2√5 is irrational.
Show Solution ▶
Assume 3 + 2√5 is rational. Then 3 + 2√5 = a/b for coprime integers a, b (b ≠ 0).
Rearranging: 2√5 = a/b − 3 = (a − 3b)/b
So √5 = (a − 3b)/(2b).
Since a, b, 3, and 2 are all integers, (a − 3b)/(2b) is rational — so √5 would be rational.
But this contradicts the fact that √5 is irrational (proved above).
Therefore 3 + 2√5 is irrational.
Rearranging: 2√5 = a/b − 3 = (a − 3b)/b
So √5 = (a − 3b)/(2b).
Since a, b, 3, and 2 are all integers, (a − 3b)/(2b) is rational — so √5 would be rational.
But this contradicts the fact that √5 is irrational (proved above).
Therefore 3 + 2√5 is irrational.
| 📝 Practice Set A — Multiple Choice Questions (8 Questions) |
Q1. The prime factorisation of 156 is:
(a) 2² × 3 × 13 (b) 2 × 3² × 13 (c) 2³ × 3 × 13 (d) 2² × 39
(a) 2² × 3 × 13 (b) 2 × 3² × 13 (c) 2³ × 3 × 13 (d) 2² × 39
Show Answer ▶
(a) 2² × 3 × 13 — 156 ÷ 2 = 78 → 78 ÷ 2 = 39 → 39 ÷ 3 = 13 → 13 is prime. So 156 = 2² × 3 × 13. Check: 4 × 3 × 13 = 4 × 39 = 156 ✓
Q2. HCF(6, 20) × LCM(6, 20) equals:
(a) 26 (b) 60 (c) 120 (d) 240
(a) 26 (b) 60 (c) 120 (d) 240
Show Answer ▶
(c) 120 — By the relation HCF × LCM = a × b = 6 × 20 = 120. (HCF = 2, LCM = 60, and 2 × 60 = 120 = 6 × 20 ✓)
Q3. Which of these is an irrational number?
(a) √9 (b) √(4/9) (c) √7 (d) √(49/16)
(a) √9 (b) √(4/9) (c) √7 (d) √(49/16)
Show Answer ▶
(c) √7 — √9 = 3 (rational); √(4/9) = 2/3 (rational); √7 is irrational (7 is prime, so √7 is irrational by Theorem 1.2); √(49/16) = 7/4 (rational).
Q4. If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) is:
(a) 150 (b) 90 (c) 300 (d) 216
(a) 150 (b) 90 (c) 300 (d) 216
Show Answer ▶
(a) 150 — LCM = (a × b) / HCF = 1800 / 12 = 150. Verify: HCF × LCM = 12 × 150 = 1800 = a × b ✓
Q5. The number 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is:
(a) Prime (b) Composite (c) Neither prime nor composite (d) 1
(a) Prime (b) Composite (c) Neither prime nor composite (d) 1
Show Answer ▶
(b) Composite — 7! + 5 = 5040 + 5 = 5045. Factor out 5: = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009. Since it has 5 as a factor (other than 1 and itself), it is composite.
Q6. LCM of 8, 9 and 25 is:
(a) 1800 (b) 3600 (c) 900 (d) 450
(a) 1800 (b) 3600 (c) 900 (d) 450
Show Answer ▶
(a) 1800 — 8 = 2³; 9 = 3²; 25 = 5². LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800. (No common prime factors between 8, 9 and 25, so LCM = their product.)
Q7. In the proof that √2 is irrational, after showing 2 | a², we concluded 2 | a. This used:
(a) The fact that 2 is even (b) Theorem: if prime p | a², then p | a (c) Long division (d) The definition of irrational numbers
(a) The fact that 2 is even (b) Theorem: if prime p | a², then p | a (c) Long division (d) The definition of irrational numbers
Show Answer ▶
(b) Theorem 1.2: if prime p divides a², then p divides a — This is the key lemma used in all irrationality proofs. It relies on the uniqueness part of the Fundamental Theorem of Arithmetic.
Q8. Which of the following cannot end with digit 0 for any natural number n?
(a) 10ⁿ (b) 30ⁿ (c) 9ⁿ (d) 20ⁿ
(a) 10ⁿ (b) 30ⁿ (c) 9ⁿ (d) 20ⁿ
Show Answer ▶
(c) 9ⁿ — 9 = 3², so 9ⁿ = 3²ⁿ. Its prime factorisation only contains 3. For a number to end in 0, it needs the prime 5 in its factorisation. Since 9ⁿ never contains 5, it can never end in 0. (10ⁿ, 30ⁿ, 20ⁿ all contain both 2 and 5, so they always end in 0.)
| 📝 Practice Set B — Short Answer Questions (6 Questions) |
Q1. Find LCM and HCF of 510 and 92. Verify LCM × HCF = product of the two numbers.
Show Answer ▶
510 = 2 × 3 × 5 × 17; 92 = 2² × 23
Common prime: 2. Smallest power: 2¹. HCF(510, 92) = 2
All primes: 2, 3, 5, 17, 23. Greatest powers: 2², 3¹, 5¹, 17¹, 23¹.
LCM(510, 92) = 4 × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920; Product = 510 × 92 = 46920 ✓
Common prime: 2. Smallest power: 2¹. HCF(510, 92) = 2
All primes: 2, 3, 5, 17, 23. Greatest powers: 2², 3¹, 5¹, 17¹, 23¹.
LCM(510, 92) = 4 × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920; Product = 510 × 92 = 46920 ✓
Q2. Prove that 1/√2 is irrational.
Show Answer ▶
Assume 1/√2 is rational. Then 1/√2 = a/b (a, b coprime integers, b ≠ 0).
Rearranging: √2 = b/a.
Since b and a are integers, b/a is rational — so √2 is rational.
But this contradicts the proven fact that √2 is irrational.
Therefore 1/√2 is irrational.
Rearranging: √2 = b/a.
Since b and a are integers, b/a is rational — so √2 is rational.
But this contradicts the proven fact that √2 is irrational.
Therefore 1/√2 is irrational.
Q3. Find LCM and HCF of 12, 15 and 21 by prime factorisation.
Show Answer ▶
12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7
Prime common to ALL three: 3. Smallest power: 3¹.
HCF(12, 15, 21) = 3
All primes involved: 2, 3, 5, 7. Greatest powers: 2², 3¹, 5¹, 7¹.
LCM(12, 15, 21) = 4 × 3 × 5 × 7 = 420
Prime common to ALL three: 3. Smallest power: 3¹.
HCF(12, 15, 21) = 3
All primes involved: 2, 3, 5, 7. Greatest powers: 2², 3¹, 5¹, 7¹.
LCM(12, 15, 21) = 4 × 3 × 5 × 7 = 420
Q4. Prove that 7√5 is irrational.
Show Answer ▶
Assume 7√5 is rational. Then 7√5 = a/b (a, b coprime integers, b ≠ 0).
Rearranging: √5 = a/(7b).
Since a, 7 and b are integers, a/(7b) is rational — so √5 is rational.
But this contradicts the proven fact that √5 is irrational.
Therefore 7√5 is irrational.
Rearranging: √5 = a/(7b).
Since a, 7 and b are integers, a/(7b) is rational — so √5 is rational.
But this contradicts the proven fact that √5 is irrational.
Therefore 7√5 is irrational.
Q5. Express 3825 as a product of its prime factors.
Show Answer ▶
3825 ÷ 3 = 1275 → 1275 ÷ 3 = 425 → 425 ÷ 5 = 85 → 85 ÷ 5 = 17 → 17 is prime.
3825 = 3² × 5² × 17
Check: 9 × 25 × 17 = 225 × 17 = 3825 ✓
3825 = 3² × 5² × 17
Check: 9 × 25 × 17 = 225 × 17 = 3825 ✓
Q6. Prove that 6 + √2 is irrational.
Show Answer ▶
Assume 6 + √2 is rational. Then 6 + √2 = a/b (a, b coprime, b ≠ 0).
Rearranging: √2 = a/b − 6 = (a − 6b)/b.
Since a, 6 and b are integers, (a − 6b)/b is rational — so √2 is rational.
But this contradicts the fact that √2 is irrational.
Therefore 6 + √2 is irrational.
Rearranging: √2 = a/b − 6 = (a − 6b)/b.
Since a, 6 and b are integers, (a − 6b)/b is rational — so √2 is rational.
But this contradicts the fact that √2 is irrational.
Therefore 6 + √2 is irrational.
| 📝 Practice Set C — Long Answer Questions (4 Questions) |
Q1. State and explain the Fundamental Theorem of Arithmetic. Give two applications with examples.
Show Answer ▶
Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Application 1 — Finding HCF and LCM: To find HCF(96, 120) and LCM(96, 120): 96 = 2⁵ × 3; 120 = 2³ × 3 × 5. HCF = 2³ × 3 = 24 (smallest powers of common factors). LCM = 2⁵ × 3 × 5 = 480 (greatest powers of all factors). Verify: 24 × 480 = 11520 = 96 × 120 ✓
Application 2 — Proving irrationality: To prove √2 is irrational, we use the uniqueness of prime factorisation. If 2 | a², then 2 must appear in the factorisation of a² — so 2 | a. This lemma drives the proof by contradiction showing √2 cannot be expressed as p/q.
Q2. Prove step-by-step that √2 is irrational. What would happen if we did not require a and b to be coprime?
Show Answer ▶
Proof: Assume √2 is rational. Then √2 = r/s for some integers r, s (s ≠ 0).
Divide by any common factor to get √2 = a/b where a and b are coprime (HCF = 1).
So b√2 = a. Squaring: 2b² = a². So 2 | a². By Theorem 1.2, 2 | a. Write a = 2c.
Substituting: 2b² = 4c² → b² = 2c². So 2 | b² → 2 | b.
Both 2 | a and 2 | b — contradicts HCF(a,b) = 1.
Therefore √2 is irrational.
Divide by any common factor to get √2 = a/b where a and b are coprime (HCF = 1).
So b√2 = a. Squaring: 2b² = a². So 2 | a². By Theorem 1.2, 2 | a. Write a = 2c.
Substituting: 2b² = 4c² → b² = 2c². So 2 | b² → 2 | b.
Both 2 | a and 2 | b — contradicts HCF(a,b) = 1.
Therefore √2 is irrational.
Why coprimeness is essential: If we didn’t require a and b to be coprime, we couldn’t derive the contradiction. For example, if a = 4, b = 2 (not coprime), both being divisible by 2 tells us nothing — it’s expected. The coprimeness condition is what makes discovering “2 divides both a and b” a genuine contradiction.
Q3. Find LCM of 17, 23 and 29. What special property do these numbers have, and how does it affect the LCM?
Show Answer ▶
17, 23 and 29 are all prime numbers.
Prime factorisations: 17 = 17¹; 23 = 23¹; 29 = 29¹
Since all three are distinct primes, there are no common factors between any two of them.
HCF(17, 23, 29) = 1 (they are pairwise coprime).
LCM = greatest power of each prime = 17¹ × 23¹ × 29¹ = 17 × 23 × 29 = 11339
Special property: When numbers are pairwise coprime (or all prime), LCM = their product. This is because they share no prime factors.
Prime factorisations: 17 = 17¹; 23 = 23¹; 29 = 29¹
Since all three are distinct primes, there are no common factors between any two of them.
HCF(17, 23, 29) = 1 (they are pairwise coprime).
LCM = greatest power of each prime = 17¹ × 23¹ × 29¹ = 17 × 23 × 29 = 11339
Special property: When numbers are pairwise coprime (or all prime), LCM = their product. This is because they share no prime factors.
Q4. What is proof by contradiction? Illustrate by proving that the sum of a rational and an irrational number is irrational. (General case)
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Proof by Contradiction: To prove proposition P is true, we assume P is false (assume its negation ¬P is true). We then use logical steps to arrive at a statement that is clearly false (a contradiction). Since logical reasoning from a true assumption cannot produce a false conclusion, our assumption ¬P must be false — therefore P must be true.
Theorem: If r is rational (r ≠ 0) and x is irrational, then r + x is irrational.
Proof: Assume r + x is rational. Say r + x = q where q is rational.
Then x = q − r. Since both q and r are rational, their difference q − r is rational.
So x is rational — but this contradicts our assumption that x is irrational.
This contradiction arose from assuming r + x is rational.
Therefore r + x is irrational.
| 📝 Practice Set D — Numerical Problems (4 Questions) |
Q1. Express 5005 as a product of prime factors.
Show Answer ▶
5005 ÷ 5 = 1001 → 1001 ÷ 7 = 143 → 143 ÷ 11 = 13 → 13 is prime.
5005 = 5 × 7 × 11 × 13
Check: 5 × 7 = 35; 35 × 11 = 385; 385 × 13 = 5005 ✓
5005 = 5 × 7 × 11 × 13
Check: 5 × 7 = 35; 35 × 11 = 385; 385 × 13 = 5005 ✓
Q2. Find HCF and LCM of 336 and 54. Verify LCM × HCF = 336 × 54.
Show Answer ▶
336 = 2⁴ × 3 × 7; 54 = 2 × 3³
Common primes: 2 and 3. Smallest powers: 2¹, 3¹.
HCF(336, 54) = 2 × 3 = 6
All primes: 2, 3, 7. Greatest powers: 2⁴, 3³, 7¹.
LCM(336, 54) = 16 × 27 × 7 = 3024
Verification: 6 × 3024 = 18144; 336 × 54 = 18144 ✓
Common primes: 2 and 3. Smallest powers: 2¹, 3¹.
HCF(336, 54) = 2 × 3 = 6
All primes: 2, 3, 7. Greatest powers: 2⁴, 3³, 7¹.
LCM(336, 54) = 16 × 27 × 7 = 3024
Verification: 6 × 3024 = 18144; 336 × 54 = 18144 ✓
Q3. Two bells ring at intervals of 15 minutes and 20 minutes respectively. If they ring together at 9:00 AM, at what time will they next ring together?
Show Answer ▶
They will ring together again after LCM(15, 20) minutes.
15 = 3 × 5; 20 = 2² × 5
LCM = 2² × 3 × 5 = 60 minutes.
Starting at 9:00 AM, they will next ring together at 10:00 AM.
15 = 3 × 5; 20 = 2² × 5
LCM = 2² × 3 × 5 = 60 minutes.
Starting at 9:00 AM, they will next ring together at 10:00 AM.
Q4. Express 7429 as a product of prime factors.
Show Answer ▶
7429 ÷ 17 = 437 → 437 ÷ 19 = 23 → 23 is prime.
7429 = 17 × 19 × 23
Check: 17 × 19 = 323; 323 × 23 = 7429 ✓
7429 = 17 × 19 × 23
Check: 17 × 19 = 323; 323 × 23 = 7429 ✓
| 💡 Important Facts to Remember |
⚡ HCF × LCM = a × b rule is ONLY for two numbers: For three or more numbers, HCF × LCM ≠ product of numbers. Don’t make this mistake in the exam!
⚡ Carl Friedrich Gauss (1777–1855): The first correct proof of the Fundamental Theorem of Arithmetic was given by Gauss in his Disquisitiones Arithmeticae. He is called the “Prince of Mathematicians” and is considered one of the three greatest mathematicians of all time (along with Archimedes and Newton).
⚡ Note on 1: The number 1 is NEITHER prime nor composite. It has exactly one factor (itself), which doesn’t fit the definition of either. The Fundamental Theorem of Arithmetic starts with composite numbers.
⚡ Every natural number n > 1 is either prime or can be expressed as a unique product of primes. This is the essence of the Fundamental Theorem. Combined with Theorem 1.2, it is the foundation for proving irrationality of √p for any prime p.
| 📚 Chapter Summary |
Chapter 1 — Real Numbers: Key Points
| Fundamental Theorem of Arithmetic Every composite number = unique product of primes (apart from order). Used to find HCF and LCM, and to prove irrationality. |
Theorem 1.2 (Key Lemma) If prime p divides a², then p divides a. This is the engine behind all irrationality proofs. Uses uniqueness of prime factorisation. |
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| HCF and LCM Formulae HCF = product of smallest powers of common primes. LCM = product of greatest powers of all primes. HCF × LCM = a × b (two numbers only!) |
Proof by Contradiction Assume opposite → derive contradiction → original must be true. Used to prove: √2, √3, √5, 5−√3, 3√2, 7√5 are all irrational. |
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| 🏆 8-Point Exam Quick-Check |
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✅ Must-Know Facts
1. Every composite number has a unique prime factorisation (apart from order). 2. HCF = smallest power of common primes; LCM = greatest power of all primes. 3. HCF × LCM = a × b (for two numbers ONLY — not three!). 4. If prime p | a², then p | a (Theorem 1.2 — used in every irrationality proof). 5. √2, √3, √5, and √p for any prime p are all irrational. |
⚠ Common Exam Traps
Trap 1: HCF × LCM = a × b is for TWO numbers only. For three, it does NOT hold. Trap 2: 1 is neither prime nor composite! Don’t include 1 as a prime factor. Trap 3: In irrationality proofs, always start by assuming the number IS rational (take the negation). Trap 4: The coprimeness of a and b is ESSENTIAL in the proof — don’t skip this step. Trap 5: A number ends in 0 only if BOTH 2 AND 5 appear in its prime factorisation. |
This comprehensive study page for Class 10 Maths Chapter 1 — Real Numbers covers all CBSE and NCERT topics including the Fundamental Theorem of Arithmetic, prime factorisation, HCF and LCM by prime factorisation method, the relationship HCF × LCM = product of two numbers, and complete proof-by-contradiction proofs that √2, √3, √5, and combinations like 5−√3 and 3√2 are irrational. Includes 14 fully solved worked examples, 4 practice sets with answers, HTML table factor tree diagram, and HCF/LCM comparison charts. Perfect for CBSE Class 10 board exam preparation. Visit School Revise for more Grade 10 NCERT Maths study materials.
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