Introduction

Chapter 1: Real Numbers

Page 1 of 5: The Fundamental Theorem of Arithmetic

Welcome to the fascinating world of Real Numbers! In Class 9, you explored different types of numbers, including the mysterious irrational numbers. Now, we'll dive deeper into the properties of positive integers, the very building blocks of arithmetic.

This chapter is built on two powerful ideas: Euclid's Division Algorithm and the Fundamental Theorem of Arithmetic. Think of them as two different lenses to view numbers. Euclid's algorithm helps us understand numbers through division, while the Fundamental Theorem of Arithmetic reveals their secrets through multiplication.

{{FORMULA: expr=HCF(a, b) × LCM(a, b) = a × b | symbols=a:first integer, b:second integer, HCF:Highest Common Factor, LCM:Least Common Multiple}}

Concept Introduction

The Fundamental Theorem of Arithmetic (FTA) is a cornerstone of number theory. It states that every composite number can be expressed as a product of prime numbers in a unique way. It's like saying every complex molecule is made of a unique combination of atoms. No matter how you break it down, you'll always end up with the same set of fundamental particles.

This "unique factorization" is more than just a mathematical curiosity. It’s the backbone of modern digital security.

Real-Life Example: Cryptography

Every time you send a WhatsApp message or make an online payment, your data is encrypted using large numbers. This security relies on a simple fact: it's easy to multiply two large prime numbers together, but it's extremely difficult to take the resulting product and figure out the original prime factors. The uniqueness guaranteed by the Fundamental Theorem of Arithmetic ensures that there is only one correct pair of prime factors, making it a secure "key" to lock your information.

Definitions & Formulas

Before we explore the theorem, let's refresh some key terms. These are the basic tools you'll use throughout this chapter.

The Logic of Unique Prime Factorization

The Fundamental Theorem of Arithmetic might sound complex, but the idea behind it is something you're already familiar with: the factor tree. Let's see how it demonstrates the theorem's core logic.

1. Start with a composite number. Let's take the number 32760, as mentioned in the NCERT text.

2. Break it down into any two factors. An obvious one is 3276 × 10.

3. Continue breaking down composite factors.

   • 10 is easy: 2 × 5. Both are prime, so we stop there.
   • 3276 is even, so it's divisible by 2: 3276 = 2 × 1638.
   • 1638 is also even: 1638 = 2 × 819.
   • 819 looks divisible by 3 (sum of digits is 18): 819 = 3 × 273.
   • 273 is also divisible by 3 (sum of digits is 12): 273 = 3 × 91.
   • 91 is a classic one: 91 = 7 × 13.

4. Collect all the prime "leaves" of the tree. If we gather all the numbers at the end of our branches, we get: 2, 5, 2, 2, 3, 3, 7, 13.

{{VISUAL: diagram: A factor tree breaking down the number 32760 into its prime factors, with branches showing divisions until only prime numbers (2, 3, 5, 7, 13) remain at the end.}}

5. Write the final product. No matter how you start factoring (e.g., 32760 = 5 × 6552), you will always end up with the same set of prime factors. Let's write them in ascending order and use exponents for repeated factors.

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13

32760 = 2³ × 3² × 5¹ × 7¹ × 13¹

This final expression is the unique prime factorization of 32760. This process leads us directly to the formal statement of the theorem.

{{KEY: type=theorem | title=The Fundamental Theorem of Arithmetic (Theorem 1.1) | text=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.}}

Solved Examples

Let's apply this theorem to solve some problems, from basic to advanced.

Example 1: Basic Prime Factorization (Easy)

Given: The number 140.

To Find: The prime factorization of 140.

Solution:

1. Start by dividing 140 by the smallest prime number, 2.

   140 ÷ 2 = 70
   

2. The result, 70, is also divisible by 2.

   70 ÷ 2 = 35
   

3. The new result, 35, is not divisible by 2 or 3. The next prime is 5.

   35 ÷ 5 = 7
   

4. The result, 7, is a prime number. Now, collect all the prime factors.

   140 = 2 × 2 × 5 × 7
   

5. Express the factorization using powers.

   140 = 2² × 5 × 7
   

Final Answer: The prime factorization of 140 is 2² × 5 × 7.

Example 2: Finding HCF and LCM of Two Numbers (Medium)

Given: Integers 510 and 92.

To Find: The HCF and LCM of 510 and 92, and verify that HCF × LCM = Product of the two numbers.

Solution:

1. Find the prime factorization of 510.

   510 = 2 × 255 = 2 × 3 × 85 = 2 × 3 × 5 × 17
   

2. Find the prime factorization of 92.

   92 = 2 × 46 = 2 × 2 × 23 = 2² × 23
   

3. To find the HCF, take the product of the smallest power of each common prime factor. The only common prime factor is 2, and its smallest power is 2¹.

   HCF(510, 92) = 2
   

4. To find the LCM, take the product of the greatest power of all prime factors involved. The factors are 2, 3, 5, 17, and 23.

   LCM(510, 92) = 2² × 3¹ × 5¹ × 17¹ × 23¹
   

   LCM(510, 92) = 4 × 3 × 5 × 17 × 23 = 23460
   

5. Now, verify the relationship. First, calculate the product of the numbers.

   Product = 510 × 92 = 46920
   

6. Next, calculate the product of the HCF and LCM.

   HCF × LCM = 2 × 23460 = 46920
   

7. Since both products are equal, the relationship is verified.

Final Answer: HCF = 2, LCM = 23460. The verification HCF × LCM = Product of numbers holds true as 46920 = 46920.

Example 3: HCF and LCM of Three Numbers (Hard)

Given: Integers 12, 15, and 21.

To Find: The HCF and LCM using the prime factorization method.

Solution:

1. Find the prime factorization of each number.

   12 = 2 × 6 = 2 × 2 × 3 = 2² × 3¹
   

   15 = 3 × 5 = 3¹ × 5¹
   

   21 = 3 × 7 = 3¹ × 7¹
   

2. For the HCF, find the common prime factors and their lowest powers. The only prime factor common to all three is 3, and its lowest power is 3¹.

   HCF(12, 15, 21) = 3
   

3. For the LCM, find all unique prime factors (2, 3, 5, 7) and their highest powers across all factorizations (2², 3¹, 5¹, 7¹).

   LCM(12, 15, 21) = 2² × 3¹ × 5¹ × 7¹
   

   LCM(12, 15, 21) = 4 × 3 × 5 × 7 = 420
   

   Important Note: The rule HCF × LCM = Product of numbers does NOT apply for three numbers.

Final Answer: HCF = 3, LCM = 420.

Example 4: Real-World Application (Tricky)

Given: A circular sports field. Priya takes 18 minutes to complete one round, while Harish takes 12 minutes for the same. They start at the same point and at the same time, and go in the same direction.

To Find: After how many minutes will they meet again at the starting point?

Solution:

1. This problem asks for the next time they will both be at the starting point. This means we are looking for the smallest number of minutes that is a multiple of both 18 and 12. This is a classic LCM problem.

2. First, find the prime factorization of 18.

   18 = 2 × 9 = 2 × 3 × 3 = 2¹ × 3²
   

3. Next, find the prime factorization of 12.

   12 = 2 × 6 = 2 × 2 × 3 = 2² × 3¹
   

4. To find the LCM, take the highest power of all prime factors involved (2 and 3). The highest power of 2 is 2² and the highest power of 3 is 3².

   LCM(18, 12) = 2² × 3²
   

5. Calculate the final value.

   LCM(18, 12) = 4 × 9 = 36
   

   This means they will both be at the starting point after 36 minutes. Priya will have completed 36 ÷ 18 = 2 rounds, and Harish will have completed 36 ÷ 12 = 3 rounds.

Final Answer: They will meet again at the starting point after 36 minutes.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Can two numbers have 15 as their HCF and 175 as their LCM? Justify your answer.

   💡 Answer: No. A key property is that the HCF of two numbers must always be a factor of their LCM. Here, 175 ÷ 15 = 11.66..., which is not an integer. Therefore, it's impossible for two numbers to have an HCF of 15 and an LCM of 175.

2. What is the smallest number that, when divided by 35, 56, and 91, leaves a remainder of 7 in each case?

   💡 Answer: First, find the LCM of 35, 56, and 91. 35 = 5 × 7 56 = 8 × 7 = 2³ × 7 91 = 7 × 13 LCM = 2³ × 5 × 7 × 13 = 8 × 5 × 91 = 40 × 91 = 3640. This is the smallest number divisible by all three. To leave a remainder of 7, we simply add 7 to the LCM. The required number is 3640 + 7 = 3647.

3. The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

   💡 Answer: We use the formula HCF × LCM = Product of the two numbers. Let the other number be x. 23 × 1449 = 161 × x x = (23 × 1449) / 161 We know 161 = 7 × 23. x = (23 × 1449) / (7 × 23) x = 1449 / 7 x = 207. The other number is 207.

Mini Cheatsheet

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The Fundamental Theorem of Arithmetic (Part 1)

The Fundamental Theorem of Arithmetic (Part 1)

Concept Introduction

Imagine you're trying to understand the DNA of numbers. Just as every living organism has a unique genetic code, every whole number has a unique "prime DNA" — a special way to break it down into building blocks called prime numbers.

When a shopkeeper arranges 24 items into boxes, they can organize them as 2 × 12, or 3 × 8, or 4 × 6, or even 2 × 2 × 2 × 3. But there's only ONE way to write 24 using prime numbers alone: 2³ × 3. No matter how you rearrange those primes, the fundamental identity remains the same.

This powerful idea — that every composite number can be expressed as a product of primes in exactly one way — is the Fundamental Theorem of Arithmetic. It's the backbone of number theory and helps us solve problems ranging from finding HCF and LCM to proving that certain numbers (like √2) are irrational. Understanding this theorem is like unlocking the secret architecture of the entire number system.

{{FORMULA: expr=n = p₁^a₁ × p₂^a₂ × p₃^a₃ × ... × pₖ^aₖ | symbols=n:composite number, p₁,p₂,...:distinct primes in ascending order, a₁,a₂,...:positive integer powers}}

Definitions & Formulas

{{KEY: type=theorem | title=The Fundamental Theorem of Arithmetic | text=Every composite number can be expressed as a product of primes in exactly ONE way, apart from the order in which the prime factors occur.}}

Understanding the Theorem: Logic & Derivation

Let's build the intuition behind this fundamental theorem through systematic reasoning:

Step 1: Every composite number can be factorized

Take any composite number, say 60. Since it's composite, it has at least one divisor other than 1 and itself. We can write:

60 = 6 × 10

Step 2: Continue factorizing until only primes remain

Now 6 and 10 are both composite, so we break them down further:

60 = (2 × 3) × (2 × 5) = 2 × 2 × 3 × 5 = 2² × 3 × 5

Eventually, we reach prime numbers that cannot be broken down further.

Step 3: The factorization is independent of the path taken

Whether you start with 60 = 4 × 15 or 60 = 12 × 5, you'll always end up with the same prime factors:

60 = 4 × 15 = (2 × 2) × (3 × 5) = 2² × 3 × 5
60 = 12 × 5 = (2 × 2 × 3) × 5 = 2² × 3 × 5

Step 4: Writing in standard form ensures uniqueness

By convention, we write primes in ascending order and use exponents for repeated factors:

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13

Step 5: The uniqueness part is crucial

The theorem states that not only CAN every composite be factorized into primes, but there is ONLY ONE such factorization (ignoring order). This means:

If n = p₁^a₁ × p₂^a₂ × ... = q₁^b₁ × q₂^b₂ × ...
Then {p₁, p₂, ...} = {q₁, q₂, ...} and a₁ = b₁, a₂ = b₂, etc.

Step 6: Applications of uniqueness

This uniqueness property is what makes HCF and LCM calculations systematic. It also helps us prove important results like the irrationality of √2, √3, etc.

{{KEY: type=insight | title=Why "Apart from Order" Matters | text=We consider 2 × 3 × 5 and 5 × 2 × 3 as the SAME factorization because multiplication is commutative. Standard form (ascending order) removes this ambiguity.}}

Solved Examples

Example 1: Basic Prime Factorization (Easy)

Given: The number 140

To Find: Express 140 as a product of its prime factors in standard form

Solution:

1. Start by dividing by the smallest prime (2):

140 ÷ 2 = 70

2. Continue dividing 70 by 2:

70 ÷ 2 = 35

3. Now 35 is not divisible by 2, try the next prime (3) — doesn't work. Try 5:

35 ÷ 5 = 7

4. 7 is already prime, so we stop. Collecting all prime factors:

140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

Final Answer: 140 = 2² × 5 × 7

Example 2: Verifying Uniqueness (Medium)

Given: The number 156

To Find: Show that different factorization paths lead to the same prime factorization

Solution:

1. Path 1: Start with 156 = 2 × 78

156 = 2 × 78 = 2 × (2 × 39) = 2 × 2 × (3 × 13) = 2² × 3 × 13

2. Path 2: Start with 156 = 4 × 39

156 = 4 × 39 = (2 × 2) × (3 × 13) = 2² × 3 × 13

3. Path 3: Start with 156 = 12 × 13

156 = 12 × 13 = (2 × 2 × 3) × 13 = 2² × 3 × 13

4. All three paths yield the identical prime factorization, confirming the theorem's uniqueness.

Final Answer: 156 = 2² × 3 × 13 (unique factorization verified)

Example 3: Large Number Factorization (Medium-Hard)

Given: The number 5005

To Find: Prime factorization in standard form

Solution:

1. Check divisibility by 5 (ends in 5):

5005 ÷ 5 = 1001

2. Factorize 1001. Try 7:

1001 ÷ 7 = 143

3. Factorize 143. Try 11:

143 ÷ 11 = 13

4. 13 is prime. Collecting all factors in ascending order:

5005 = 5 × 7 × 11 × 13

Final Answer: 5005 = 5 × 7 × 11 × 13

Example 4: Application to Ending Digits (Tricky)

Given: Numbers of the form 4ⁿ where n is a natural number

To Find: Can 4ⁿ ever end with the digit 0 for any value of n?

Solution:

1. For a number to end in 0, it must be divisible by 10, which means divisible by both 2 AND 5.

10 = 2 × 5

2. Write 4ⁿ in its prime factorization:

4ⁿ = (2²)ⁿ = 2^(2n)

3. By the Fundamental Theorem of Arithmetic, this factorization is UNIQUE. The only prime factor is 2.

Prime factors of 4ⁿ = {2} only

4. Since 5 is NOT in the prime factorization of 4ⁿ, it cannot be divisible by 5, hence cannot be divisible by 10.

4ⁿ cannot contain factor 5 → cannot end in 0

Final Answer: No, 4ⁿ can never end with digit 0 for any natural number n

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If a number n has exactly 3 prime factors (all distinct), what is the minimum number of divisors n can have?

💡 Answer: If n = p × q × r (three distinct primes), its divisors are: 1, p, q, r, pq, pr, qr, pqr. That's 8 divisors minimum. For example, 30 = 2 × 3 × 5 has exactly 8 divisors.

Q2: Can two different composite numbers have the exact same prime factorization?

💡 Answer: No! The Fundamental Theorem of Arithmetic guarantees uniqueness. If two numbers have identical prime factorizations, they ARE the same number. This is the whole point of "unique" factorization.

Q3: The prime factorization of a number is 2³ × 3² × 5. Without calculating, determine if this number is a perfect square.

💡 Answer: No, it is NOT a perfect square. For a number to be a perfect square, ALL exponents in its prime factorization must be even. Here, the exponents are 3, 2, and 1 — both 3 and 1 are odd, so the number cannot be a perfect square.

Mini Cheatsheet

Next: We'll explore how to use the Fundamental Theorem to systematically find HCF and LCM through prime factorization, and verify the important relationship: HCF(a,b) × LCM(a,b) = a × b.

The Fundamental Theorem of Arithmetic (Part 2)

The Fundamental Theorem of Arithmetic (Part 2)

Concept Introduction

In the previous section, we discovered that every composite number can be uniquely expressed as a product of prime numbers. But why does this matter in real life?

Imagine you're organizing a school assembly with 96 students and 404 chairs. You want to arrange them in rows with the same number of students and chairs per row. What is the maximum number of rows possible? This is precisely where finding the Highest Common Factor (HCF) becomes essential.

Similarly, if two bells ring at intervals of 6 minutes and 20 minutes respectively, when will they ring together again? This requires finding the Least Common Multiple (LCM). The Fundamental Theorem of Arithmetic provides us with a powerful method—prime factorization—to compute both HCF and LCM efficiently and accurately.

Today, we'll master this technique and apply it to solve real-world problems involving divisibility and common multiples.

{{FORMULA: expr=HCF(a, b) × LCM(a, b) = a × b | symbols=a:first positive integer, b:second positive integer, HCF:highest common factor, LCM:least common multiple}}

Definitions & Formulas

{{KEY: type=formula | title=The Golden Relationship | text=For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b. This relationship ONLY holds for exactly TWO numbers, not three or more.}}

Method: Prime Factorization for HCF and LCM

Step 1: Find Prime Factorization

Express each number as a product of powers of primes in ascending order.

For example, for 72 and 120:

72 = 2³ × 3²

120 = 2³ × 3 × 5

Step 2: Identify All Prime Factors

List all distinct primes appearing in ANY of the numbers: 2, 3, 5

Step 3: Computing HCF

HCF = Product of the SMALLEST power of each COMMON prime factor

For 72 and 120, common primes are 2 and 3:

HCF(72, 120) = 2³ × 3¹ = 8 × 3 = 24

{{KEY: type=concept | title=HCF Logic | text=HCF contains only those primes that appear in ALL numbers. The power is the MINIMUM of all occurrences.}}

Step 4: Computing LCM

LCM = Product of the GREATEST power of each prime factor (whether common or not)

For 72 and 120, all primes are 2, 3, 5:

LCM(72, 120) = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360

Step 5: Verification Using the Golden Relationship

Always verify:

HCF × LCM = 24 × 360 = 8640

72 × 120 = 8640 ✓

Step 6: Alternative Path (Finding LCM from HCF)

If HCF is already known:

LCM(a, b) = (a × b) / HCF(a, b)

Solved Examples

Example 1: Basic Application (Easy)

Given: Numbers are 18 and 48

To Find: HCF and LCM using prime factorization

Solution:

1. Find the prime factorization of each number.

18 = 2 × 3²

48 = 2⁴ × 3

2. For HCF, take the smallest power of common primes (2 and 3).

HCF(18, 48) = 2¹ × 3¹ = 6

3. For LCM, take the greatest power of all primes appearing.

LCM(18, 48) = 2⁴ × 3² = 16 × 9 = 144

4. Verify using the relationship.

HCF × LCM = 6 × 144 = 864

18 × 48 = 864 ✓

Final Answer: HCF = 6, LCM = 144

Example 2: Three Numbers (Medium)

Given: Numbers are 12, 15, and 21

To Find: HCF and LCM using prime factorization

Solution:

1. Find prime factorization of each number.

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

2. Identify the only common prime in all three: 3.

HCF(12, 15, 21) = 3¹ = 3

3. For LCM, take greatest power of ALL primes (2, 3, 5, 7).

LCM(12, 15, 21) = 2² × 3 × 5 × 7

= 4 × 3 × 5 × 7 = 420

4. Notice that for three numbers, the product relationship does NOT hold.

12 × 15 × 21 = 3780

HCF × LCM = 3 × 420 = 1260 ≠ 3780

Final Answer: HCF = 3, LCM = 420

Example 3: Finding One from the Other (Hard)

Given: HCF(306, 657) = 9

To Find: LCM(306, 657)

Solution:

1. Use the golden relationship for two numbers.

HCF(a, b) × LCM(a, b) = a × b

2. Substitute the known values.

9 × LCM(306, 657) = 306 × 657

3. Calculate the product on the right side.

306 × 657 = 201,042

4. Solve for LCM.

LCM(306, 657) = 201,042 / 9 = 22,338

Final Answer: LCM = 22,338

Example 4: Real-Life Application (Tricky)

Given: Two bells ring at intervals of 18 minutes and 12 minutes. Both ring together at 9:00 AM.

To Find: When will they ring together next?

Solution:

1. The bells ring together after a time interval equal to their LCM.

18 = 2 × 3²

12 = 2² × 3

2. Calculate LCM using greatest powers of all primes.

LCM(18, 12) = 2² × 3² = 4 × 9 = 36

3. The bells ring together every 36 minutes.

4. Add 36 minutes to 9:00 AM.

9:00 AM + 36 minutes = 9:36 AM

Final Answer: The bells will ring together next at 9:36 AM

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Question 1: If HCF(a, b) = 12 and LCM(a, b) = 180, and one number is 36, find the other number.

💡 Answer: Using a × b = HCF × LCM, we get 36 × b = 12 × 180 = 2160. Therefore b = 2160 / 36 = 60.

Question 2: Can the LCM of two numbers be smaller than both numbers? Justify.

💡 Answer: No. LCM is a MULTIPLE of both numbers, so it must be at least as large as the larger number. The minimum LCM is the larger number itself (when one divides the other).

Question 3: Three numbers have HCF = 5 and their product is 1000. If two of the numbers are 10 and 20, find the third number.

💡 Answer: Let the third number be x. Product = 10 × 20 × x = 1000, so x = 1000/200 = 5. Check: Prime factorizations are 2×5, 2²×5, and 5. HCF = 5 ✓

Mini Cheatsheet

Quick Recall: For HCF think "Common, Smallest" — for LCM think "All, Greatest". This mental hook will save you from the most common errors in exams!

Revisiting Irrational Numbers

Revisiting Irrational Numbers

Concept Introduction

Irrational numbers are numbers that cannot be expressed as a fraction p/q where p and q are integers and q ≠ 0. While we learned about their existence and properties in Class IX, we never formally proved why certain numbers are irrational.

Understanding irrationality has profound real-world implications. Consider GPS navigation systems: they rely on calculating distances using the Pythagorean theorem, which frequently involves √2, √3, and other irrational numbers. These calculations can never be exact — engineers must decide how many decimal places are "good enough" for accurate positioning. A GPS error of just 10 meters could mean the difference between directing you to the correct street or a parallel one!

In this section, we'll rigorously prove that numbers like √2, √3, √5, and generally √p (where p is prime) are irrational. Our primary weapon will be proof by contradiction — assuming the opposite of what we want to prove, then showing this leads to an impossibility.

{{FORMULA: expr=If √p is rational, then √p = a/b where a,b are coprime | symbols=p:prime number, a:numerator, b:denominator (b≠0), coprime:HCF(a,b)=1}}

Definitions & Formulas

Key Theorem: Theorem 1.2 (Foundation for Our Proofs)

Before proving √2 is irrational, we need a critical result:

{{KEY: type=theorem | title=Theorem 1.2 | text=Let p be a prime number. If p divides a², then p must divide a, where a is a positive integer.}}

Proof Logic (Based on Fundamental Theorem of Arithmetic):

1. Express a in prime factorization:

a = p₁ × p₂ × ... × pₙ

2. Square both sides to get a²:

a² = (p₁ × p₂ × ... × pₙ) × (p₁ × p₂ × ... × pₙ) = p₁² × p₂² × ... × pₙ²

3. Given: p divides a². By the Fundamental Theorem of Arithmetic, every integer has a unique prime factorization.

4. Conclusion: Since p divides a², p must appear in the prime factorization of a². Due to uniqueness, p must be one of {p₁, p₂, ..., pₙ}.

5. Therefore: If p is in the factorization of a², it must also be in the factorization of a. Hence, p divides a.

This theorem is our key tool. Let's apply it!

Proving √2 is Irrational

The Proof by Contradiction Method

Strategy: Assume √2 is rational. Show this assumption creates an impossible situation (a contradiction). Therefore, our assumption must be false.

Step-by-Step Proof:

1. Assume the opposite: Suppose √2 is rational.

√2 = r/s where r, s ∈ ℤ and s ≠ 0

2. Simplify to coprime form: If r and s share common factors, divide them out.

√2 = a/b where HCF(a, b) = 1

3. Square both sides:

2 = a²/b²

2b² = a²

4. Analyze divisibility: From 2b² = a², we see 2 divides a². By Theorem 1.2, 2 must divide a.

a = 2c for some integer c

5. Substitute back:

2b² = (2c)²

2b² = 4c²

b² = 2c²

6. Second divisibility: From b² = 2c², we see 2 divides b². By Theorem 1.2 again, 2 must divide b.

7. The contradiction: Both a and b are divisible by 2. This contradicts our assumption that HCF(a, b) = 1.

8. Conclusion: Our initial assumption (√2 is rational) must be false.

Therefore, √2 is irrational

{{KEY: type=concept | title=The Power of Contradiction | text=We don't prove √2 is irrational directly. Instead, we prove that assuming it's rational creates an impossible situation where two coprime numbers share a factor.}}

Solved Examples

Example 1: Prove that √3 is Irrational (NCERT Example 5)

Given: We need to prove √3 cannot be expressed as a/b where a, b are coprime integers.

To Find: Proof that √3 is irrational.

Solution:

1. Assume the opposite: Suppose √3 is rational in simplest form.

√3 = a/b where HCF(a, b) = 1

2. Square both sides:

3 = a²/b²

3b² = a²

3. Apply Theorem 1.2 with prime p = 3: Since 3 divides a², then 3 divides a.

a = 3c for some integer c

4. Substitute a = 3c back:

3b² = (3c)² = 9c²

b² = 3c²

5. Second application: Since 3 divides b², then 3 divides b.

6. Contradiction: Both a and b are divisible by 3, contradicting HCF(a, b) = 1.

Final Answer: √3 is irrational ✓

Example 2: Prove that √5 is Irrational

Given: We need to establish irrationality of √5.

To Find: Proof using contradiction.

Solution:

1. Assumption: Let √5 be rational.

√5 = a/b where HCF(a, b) = 1

2. Square and rearrange:

5b² = a²

3. Divisibility check: 5 divides a², so 5 divides a (Theorem 1.2).

a = 5c

4. Substitute:

5b² = 25c²

b² = 5c²

5. Second divisibility: 5 divides b², so 5 divides b.

6. Contradiction reached: Both a and b divisible by 5, but we said HCF(a, b) = 1.

Final Answer: √5 is irrational ✓

Example 3: Show that 5 − √3 is Irrational (NCERT Example 6)

Given: A difference between a rational (5) and irrational (√3) number.

To Find: Prove 5 − √3 is irrational.

Solution:

1. Assume the opposite: Suppose 5 − √3 is rational.

5 − √3 = a/b where a, b ∈ ℤ, b ≠ 0

2. Rearrange to isolate √3:

5 − a/b = √3

(5b − a)/b = √3

3. Analyze the result: Since a, b, and 5 are all integers, (5b − a)/b is rational.

√3 = (5b − a)/b = rational number

4. Contradiction: We know √3 is irrational (proved earlier), but our assumption forces it to be rational.

Final Answer: 5 − √3 is irrational ✓

Example 4: Prove that 3√2 is Irrational (NCERT Example 7)

Given: Product of rational 3 and irrational √2.

To Find: Prove 3√2 is irrational.

Solution:

1. Assume the opposite: Suppose 3√2 is rational.

3√2 = a/b where a, b ∈ ℤ, b ≠ 0

2. Isolate √2:

√2 = a/(3b)

3. Analyze: Since 3, a, and b are integers, a/(3b) is rational.

√2 = a/(3b) = rational number

4. Contradiction: √2 is irrational (proven earlier), but we've shown it equals a rational number.

Final Answer: 3√2 is irrational ✓

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: If √2 + √3 were rational, what contradiction would arise? (Hint: Square both sides)

💡 Answer: Assume √2 + √3 = r (rational). Squaring: 2 + 2√6 + 3 = r² → √6 = (r² − 5)/2. Since r is rational, (r² − 5)/2 is rational. But √6 is irrational (can be proved similarly), creating a contradiction.

Q2: Can the sum of two irrational numbers ever be rational? Give an example.

💡 Answer: Yes! Example: √2 + (−√2) = 0, which is rational. Another: (3 + √5) + (2 − √5) = 5, which is rational. The key: the irrational parts must cancel out.

Q3: Prove that 1/√2 is irrational without using proof by contradiction directly on 1/√2.

💡 Answer: We know √2 is irrational. If 1/√2 were rational, say 1/√2 = p/q, then √2 = q/p would be rational (reciprocal of rational is rational). This contradicts √2 being irrational. Therefore, 1/√2 is irrational.

Mini Cheatsheet

Practice & Exercises

EXERCISE 1.1

Concept Introduction

The Fundamental Theorem of Arithmetic is the backbone of understanding number properties in mathematics. It states that every composite number can be expressed as a unique product of prime numbers, irrespective of the order in which they appear. This theorem has powerful real-life applications.

Consider a factory production line where machines work in cycles. Machine A completes a cycle every 12 minutes, Machine B every 18 minutes, and Machine C every 24 minutes. To find when all three machines align at the starting point again, we need the LCM (Least Common Multiple). To determine the largest gear that can synchronize all three, we use the HCF (Highest Common Factor). Both concepts rely entirely on prime factorization—breaking down numbers into their prime building blocks. Understanding how to factorize numbers and apply HCF/LCM formulas helps solve scheduling problems, gear ratios, packaging optimization, and even musical rhythm patterns.

{{FORMULA: expr=n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ | symbols=n:composite number, pᵢ:prime factors, aᵢ:powers of primes}}

Definitions & Formulas

{{KEY: type=theorem | title=Critical Relationship | text=For ANY two numbers: HCF × LCM = Product of the two numbers. This does NOT hold for three or more numbers!}}

Logical Foundation: Prime Factorization Method

Understanding the step-by-step process of finding HCF and LCM through prime factorization:

1. Express each number as a product of prime factors

Write each number in the form p₁^a₁ × p₂^a₂ × p₃^a₃... where p represents prime numbers and a represents their powers.

2. Identify all prime factors present

List all unique prime numbers that appear in ANY of the numbers being analyzed.

3. For HCF: Take minimum powers

For each common prime factor (appearing in ALL numbers), select the smallest exponent. Multiply these together.

HCF = product of (common primes)^(minimum power)

4. For LCM: Take maximum powers

For each prime factor (appearing in ANY number), select the largest exponent. Multiply these together.

LCM = product of (all primes)^(maximum power)

5. Verify using the fundamental property

For two numbers, check: HCF × LCM = a × b. This confirms calculation accuracy.

6. For three or more numbers

Apply the same logic, but remember: the product property does NOT apply to 3+ numbers. HCF(a,b,c) × LCM(a,b,c) ≠ a × b × c

Solved Examples

Example 1: Prime Factorization (Easy)

Given: Number 140

To Find: Express as a product of prime factors

Solution:

1. Start dividing by the smallest prime number 2.

140 ÷ 2 = 70

2. Continue dividing by 2.

70 ÷ 2 = 35

3. Now 35 is odd, so try next prime 5.

35 ÷ 5 = 7

4. 7 is itself prime, so we stop.

140 = 2 × 2 × 5 × 7 = 2² × 5¹ × 7¹

Final Answer: 140 = 2² × 5 × 7

Example 2: Finding HCF and LCM (Medium)

Given: Two numbers 26 and 91

To Find: HCF and LCM, and verify HCF × LCM = product of numbers

Solution:

1. Prime factorize both numbers.

26 = 2 × 13
91 = 7 × 13

2. Identify common and all prime factors. Common: 13. All: 2, 7, 13.

3. For HCF, take minimum power of common primes.

HCF = 13¹ = 13

4. For LCM, take maximum power of all primes.

LCM = 2¹ × 7¹ × 13¹ = 182

5. Verify the fundamental property.

HCF × LCM = 13 × 182 = 2366
26 × 91 = 2366

Final Answer: HCF = 13, LCM = 182 ✓ Verified

Example 3: Three Numbers HCF and LCM (Hard)

Given: Three numbers 12, 15, and 21

To Find: HCF and LCM using prime factorization

Solution:

1. Express each number as product of primes.

12 = 2² × 3¹
15 = 3¹ × 5¹
21 = 3¹ × 7¹

2. Identify all primes appearing: 2, 3, 5, 7.

3. For HCF, take minimum power of common primes. Only 3 appears in all three.

HCF = 3¹ = 3

4. For LCM, take maximum power of each prime across all numbers.

LCM = 2² × 3¹ × 5¹ × 7¹

5. Calculate the LCM value.

LCM = 4 × 3 × 5 × 7 = 420

Final Answer: HCF = 3, LCM = 420

Example 4: Application Problem (Tricky)

Given: HCF(306, 657) = 9

To Find: LCM(306, 657)

Solution:

1. Use the fundamental property for two numbers.

HCF × LCM = Product of the two numbers

2. Substitute known values.

9 × LCM = 306 × 657

3. Calculate the product on the right side.

306 × 657 = 201,042

4. Solve for LCM by division.

LCM = 201,042 ÷ 9 = 22,338

Final Answer: LCM(306, 657) = 22,338

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Question 1: Number Ending Analysis

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

💡 Answer: No. For a number to end in 0, it must have both 2 and 5 as prime factors. Since 6 = 2 × 3, any power of 6 equals 6ⁿ = 2ⁿ × 3ⁿ. There is no factor of 5, so 6ⁿ can never end in 0. It will always end in 6.

Question 2: Composite Number Proof

Explain why 7 × 11 × 13 + 13 is a composite number.

💡 Answer: Factor out 13 from both terms: 7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78 = 13 × 2 × 39. Since the number has factors other than 1 and itself (specifically 13, 2, 3, etc.), it is composite.

Question 3: Meeting Point Problem

Sonia takes 18 minutes to complete one round of a circular path; Ravi takes 12 minutes. Starting together, after how many minutes will they meet again at the starting point?

💡 Answer: Find LCM(18, 12). Prime factorization: 18 = 2 × 3², 12 = 2² × 3. LCM = 2² × 3² = 4 × 9 = 36 minutes. They will meet at the starting point after 36 minutes.

Mini Cheatsheet: Quick Revision Table

{{KEY: type=warning | title=Three-Number Trap | text=The property HCF × LCM = product works ONLY for two numbers. For three or more numbers, this equation does NOT hold. Always factorize separately!}}

Practice Makes Perfect: Work through all seven exercise problems systematically. Start with prime factorization, then tackle HCF/LCM pairs, verify using the fundamental property, and finally apply concepts to real-world scenarios. Mastering these techniques now builds the foundation for irrational numbers, polynomials, and advanced algebra!