📚 Class 10 NCERT Mathematics

Chapter 1: Real Numbers - Complete Study Guide

📝 Important Notes from Chapter 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.

Example: 140 = 2² × 5 × 7 (Unique factorization)

📐 HCF (Highest Common Factor)

Also called GCD (Greatest Common Divisor)
For two numbers: HCF × LCM = Product of two numbers
For three numbers: Use prime factorization - take lowest power of common primes

📊 LCM (Least Common Multiple)

Smallest number divisible by all given numbers
For prime factorization: take highest power of all primes present
For two numbers: LCM = (a × b) / HCF(a, b)

⚠️ Important Remark

For THREE numbers:

HCF(p,q,r) × LCM(p,q,r) ≠ p × q × r

But these formulas work:

LCM(p,q,r) = (p·q·r·HCF(p,q,r)) / (HCF(p,q)·HCF(q,r)·HCF(p,r))

🔍 Prime Divisibility Property

If p is a prime and p divides a², then p divides a, where a is a positive integer.

This is crucial for proving irrationality!

🔢 Irrational Numbers

Cannot be written as p/q where p, q are integers and q ≠ 0
√2, √3, √5, √p (where p is prime) are all irrational
Sum/difference of rational and irrational is irrational
Product of non-zero rational and irrational is irrational

🎯 Proving Irrationality - Method

Assume the number is rational: = a/b (coprime)
Square both sides (for square roots)
Show that a² is divisible by prime ⇒ a is divisible by prime
Show a and b have common factor - CONTRADICTION!
Hence, the number is irrational

💡 Composite Numbers Trick

To prove expressions like 7×11×13 + 13 are composite:

Factor out common terms
7×11×13 + 13 = 13(7×11 + 1) = 13 × 78
This shows it has factors other than 1 and itself

📖 Exercise 1.1 - Real Numbers

Express each number as a product of its prime factors:

(i) 140

A 2 × 5 × 7

B 2² × 5 × 7

C 2 × 5² × 7

D 2³ × 5 × 7

✅ Solution:

Step 1: Divide 140 by smallest prime 2: 140 ÷ 2 = 70

Step 2: 70 ÷ 2 = 35

Step 3: 35 ÷ 5 = 7

Step 4: 7 ÷ 7 = 1

Answer: 140 = 2² × 5 × 7

(ii) 156

A 2² × 3 × 13

B 2² × 3 × 13

C 2 × 3² × 13

D 2³ × 3 × 13

✅ Solution:

Step 1: 156 ÷ 2 = 78

Step 2: 78 ÷ 2 = 39

Step 3: 39 ÷ 3 = 13

Step 4: 13 ÷ 13 = 1

Answer: 156 = 2² × 3 × 13

(iii) 3825

A 3² × 5² × 17

B 3² × 5² × 17

C 3 × 5³ × 17

D 3³ × 5² × 17

✅ Solution:

Step 1: 3825 ÷ 3 = 1275

Step 2: 1275 ÷ 3 = 425

Step 3: 425 ÷ 5 = 85

Step 4: 85 ÷ 5 = 17

Step 5: 17 ÷ 17 = 1

Answer: 3825 = 3² × 5² × 17

(iv) 5005

A 5 × 7 × 11 × 13

B 5 × 7 × 11 × 13

C 5 × 7 × 11 × 17

D 5 × 7 × 13 × 11

✅ Solution:

Step 1: 5005 ÷ 5 = 1001

Step 2: 1001 ÷ 7 = 143

Step 3: 143 ÷ 11 = 13

Step 4: 13 ÷ 13 = 1

Answer: 5005 = 5 × 7 × 11 × 13

(v) 7429

A 17 × 19 × 23

B 17 × 19 × 23

C 17 × 23 × 19

D 19 × 23 × 17

✅ Solution:

Step 1: 7429 ÷ 17 = 437

Step 2: 437 ÷ 19 = 23

Step 3: 23 ÷ 23 = 1

Answer: 7429 = 17 × 19 × 23

Find the LCM and HCF of the following pairs and verify LCM × HCF = Product of two numbers:

(i) 26 and 91

A HCF=13, LCM=182, Product=2366

B HCF=13, LCM=182, Product=2366

C HCF=26, LCM=91, Product=2366

D HCF=1, LCM=2366, Product=2366

✅ Solution:

Step 1: Prime factorization: 26 = 2 × 13, 91 = 7 × 13

Step 2: HCF = 13 (common factor)

Step 3: LCM = 2 × 7 × 13 = 182

Step 4: Verification: LCM × HCF = 182 × 13 = 2366

Step 5: Product = 26 × 91 = 2366 ✓

(ii) 510 and 92

A HCF=2, LCM=23460, Product=46920

B HCF=2, LCM=23460, Product=46920

C HCF=4, LCM=11730, Product=46920

D HCF=1, LCM=46920, Product=46920

✅ Solution:

Step 1: 510 = 2 × 3 × 5 × 17, 92 = 2² × 23

Step 2: HCF = 2 (lowest power of common prime)

Step 3: LCM = 2² × 3 × 5 × 17 × 23 = 23460

Step 4: Verification: 23460 × 2 = 46920

Step 5: Product = 510 × 92 = 46920 ✓

(iii) 336 and 54

A HCF=6, LCM=3024, Product=18144

B HCF=6, LCM=3024, Product=18144

C HCF=12, LCM=1512, Product=18144

D HCF=3, LCM=6048, Product=18144

✅ Solution:

Step 1: 336 = 2⁴ × 3 × 7, 54 = 2 × 3³

Step 2: HCF = 2 × 3 = 6

Step 3: LCM = 2⁴ × 3³ × 7 = 3024

Step 4: Verification: 3024 × 6 = 18144

Step 5: Product = 336 × 54 = 18144 ✓

Find LCM and HCF by prime factorization method:

(i) 12, 15 and 21

A HCF=1, LCM=420

B HCF=3, LCM=420

C HCF=3, LCM=210

D HCF=1, LCM=210

✅ Solution:

Step 1: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7

Step 2: HCF = 3 (only common prime with lowest power)

Step 3: LCM = 2² × 3 × 5 × 7 = 420

Note: For 3 numbers, HCF × LCM ≠ Product!

(ii) 17, 23 and 29

A HCF=1, LCM=11339

B HCF=1, LCM=11339

C HCF=1, LCM=11349

D HCF=29, LCM=391

✅ Solution:

Step 1: 17 = 17, 23 = 23, 29 = 29 (all prime!)

Step 2: HCF = 1 (no common factors)

Step 3: LCM = 17 × 23 × 29 = 11339

Note: They are co-prime to each other!

(iii) 8, 9 and 25

A HCF=1, LCM=1800

B HCF=1, LCM=1800

C HCF=1, LCM=180

D HCF=1, LCM=3600

✅ Solution:

Step 1: 8 = 2³, 9 = 3², 25 = 5²

Step 2: HCF = 1 (no common prime factors)

Step 3: LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800

Note: Since HCF=1, LCM = Product of all three!

Given HCF(306, 657) = 9, find LCM(306, 657).

A 22338

B 22338

C 201042

D 5913

✅ Solution:

Formula: LCM × HCF = Product of two numbers

Step 1: Product = 306 × 657 = 201042

Step 2: LCM = Product / HCF = 201042 / 9

Step 3: LCM = 22338

Verification: 22338 × 9 = 201042 ✓

Check whether 6ⁿ can end with digit 0 for any natural number n.

A Yes, for n = 5

B No, never

C Yes, for all even n

D Yes, for n = 10

✅ Solution:

Key Concept: A number ends with 0 if divisible by both 2 and 5

Step 1: 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

Step 2: Prime factorization contains only 2 and 3

Step 3: There is NO factor of 5 in 6ⁿ

Conclusion: 6ⁿ can NEVER end with digit 0 for any natural number n

Explain why 7×11×13 + 13 and 7×6×5×4×3×2×1 + 5 are composite numbers.

A They are even numbers

B They can be factored into smaller integers > 1

C They are divisible by 10

D They are prime numbers

✅ Solution:

Part 1: 7×11×13 + 13 = 13(7×11 + 1) = 13 × 78

This has factors 13 and 78 (both > 1), so it's composite.

Part 2: 7×6×5×4×3×2×1 + 5 = 5(7×6×4×3×2×1 + 1) = 5 × 1009

This has factors 5 and 1009 (both > 1), so it's composite.

Key Idea: Factor out the common term to show it has factors other than 1 and itself.

Sonia takes 18 min, Ravi takes 12 min for one round. After how many minutes will they meet again at the starting point?

A 30 minutes

B 24 minutes

C 36 minutes

D 6 minutes

✅ Solution:

Concept: They meet when both complete whole number of rounds

Step 1: Find LCM of 18 and 12

Step 2: 18 = 2 × 3², 12 = 2² × 3

Step 3: LCM = 2² × 3² = 4 × 9 = 36

Answer: They will meet after 36 minutes

Check: Sonia completes 2 rounds, Ravi completes 3 rounds

📖 Exercise 1.2 - Irrational Numbers

Prove that √5 is irrational.

A √5 = 2.236, which is terminating

B Assume √5 = a/b, show contradiction that 5 divides both a and b

C √5 cannot be measured on number line

D 5 is not a perfect square

✅ Complete Proof:

Step 1: Assume √5 is rational, so √5 = a/b where a, b are coprime integers, b ≠ 0

Step 2: Squaring both sides: 5 = a²/b² ⇒ a² = 5b²

Step 3: So 5 divides a², which means 5 divides a (by prime divisibility)

Step 4: Let a = 5c for some integer c

Step 5: Substituting: (5c)² = 5b² ⇒ 25c² = 5b² ⇒ 5c² = b²

Step 6: So 5 divides b², which means 5 divides b

Step 7: Therefore 5 divides both a and b

Step 8: This CONTRADICTS that a and b are coprime!

Conclusion: √5 is irrational. ∎

Prove that 3 + 2√5 is irrational.

A It has √5 which is irrational

B Assume it's rational, isolate √5, show contradiction

C Sum of rational and irrational is always irrational

D It cannot be written as p/q directly

✅ Complete Proof:

Step 1: Assume 3 + 2√5 is rational = a/b (coprime)

Step 2: 2√5 = a/b - 3 = (a - 3b)/b

Step 3: √5 = (a - 3b)/(2b)

Step 4: Since a, b are integers, (a-3b)/(2b) is rational

Step 5: This means √5 is rational

Step 6: But we proved √5 is irrational!

Conclusion: CONTRADICTION! Hence 3 + 2√5 is irrational. ∎

Prove that the following are irrationals:

(i) 1/√2

A It's a fraction so it's rational

B Rationalize and show contradiction

C It's less than 1 so irrational

D √2 is in denominator

✅ Solution:

Step 1: Assume 1/√2 is rational = a/b (coprime)

Step 2: √2 = b/a

Step 3: Since a, b are integers, b/a is rational

Step 4: So √2 is rational - CONTRADICTION!

Conclusion: 1/√2 is irrational. ∎

(ii) 7√5

A Product of integers is rational

B Assume rational, isolate √5, contradiction

C 7 is rational, √5 is irrational, product is irrational

D It's greater than 7

✅ Solution:

Step 1: Assume 7√5 is rational = a/b (coprime)

Step 2: √5 = a/(7b)

Step 3: Since a, b are integers, a/(7b) is rational

Step 4: So √5 is rational - CONTRADICTION!

Conclusion: 7√5 is irrational. ∎

(iii) 6 + √2

A Sum of integers is rational

B Assume rational, isolate √2, contradiction

C √2 ≈ 1.414, so 6 + 1.414 = 7.414

D It has a radical sign

✅ Solution:

Step 1: Assume 6 + √2 is rational = a/b (coprime)

Step 2: √2 = a/b - 6 = (a - 6b)/b

Step 3: Since a, b are integers, (a-6b)/b is rational

Step 4: So √2 is rational - CONTRADICTION!

Conclusion: 6 + √2 is irrational. ∎

🧮 Interactive Calculators

🔢 Prime Factorization

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📐 HCF Calculator

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📊 LCM Calculator

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✅ HCF & LCM Verifier

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⏱️ LCM Word Problem Solver

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🔍 Composite Number Checker

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