Gonit Vigyan

 

🏆 Exploring Real Numbers: A Journey into the World of Mathematics

Mathematics is full of surprises! In Class IX, you took your first step into the fascinating world of real numbers, discovering irrational numbers along the way. Now, as we move forward, we will explore two powerful mathematical tools that shape our understanding of numbers:

🔹 Euclid’s Division Algorithm – a simple yet effective method to find the HCF (Highest Common Factor) of two numbers. It also helps us understand how numbers divide each other.

🔹 Fundamental Theorem of Arithmetic – a key principle that tells us every composite number can be uniquely expressed as a product of prime numbers. This theorem plays a crucial role in proving the irrationality of numbers like √2, √3, and √5. It also helps us determine whether a rational number has a terminating or repeating decimal expansion.

Both of these concepts are easy to understand yet hold deep significance in mathematics. Through this journey, we will uncover their applications, making math more interesting and meaningful.

Let’s dive in and explore the magic of real numbers! 🚀✨ 

📖 Notes on Euclid’s Division Lemma & Algorithm

🔹 Understanding Euclid’s Division Lemma

Euclid’s Division Lemma is a simple yet powerful concept in number theory. It states that for any two positive integers a and b (where a > b), there exist unique whole numbers q (quotient) and r (remainder) such that:

$$a = bq + r, \quad \text{where } 0 \leq r < b$$

This lemma is essentially a mathematical way of representing the long division process you’ve been using for years!

💡 Example:

Let’s take some numbers and apply this concept:

1️⃣ 17 ÷ 6 → Quotient = 2, Remainder = 5
👉 $17 = 6 \times 2 + 5$

2️⃣ 5 ÷ 12 → Quotient = 0, Remainder = 5
👉 $5 = 12 \times 0 + 5$

3️⃣ 20 ÷ 4 → Quotient = 5, Remainder = 0
👉 $20 = 4 \times 5 + 0$

Did you notice? The remainder is always smaller than the divisor! That’s the key rule in the division lemma.

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🔹 Euclid’s Division Algorithm: Finding HCF

Euclid’s Division Algorithm is a step-by-step method to find the Highest Common Factor (HCF) of two numbers using the division lemma.

🛠️ Steps to Find HCF Using Euclid’s Algorithm:

1️⃣ Divide the larger number (c) by the smaller number (d) to get a quotient q and remainder r.

$$c = d \times q + r$$

2️⃣ If r = 0, then d is the HCF.
3️⃣ If r ≠ 0, repeat the process by dividing d by r until the remainder becomes 0.
4️⃣ The last non-zero divisor is the HCF!

💡 Example: Find HCF of 455 and 42

1️⃣ 455 ÷ 42 → Quotient = 10, Remainder = 35
👉 $455 = 42 \times 10 + 35$

2️⃣ 42 ÷ 35 → Quotient = 1, Remainder = 7
👉 $42 = 35 \times 1 + 7$

3️⃣ 35 ÷ 7 → Quotient = 5, Remainder = 0
👉 $35 = 7 \times 5 + 0$

🎯 Since the remainder is 0, the HCF is 7!

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🔹 Applications of Euclid’s Algorithm

Euclid’s algorithm is more than just a way to find HCF. It has deeper applications in number theory and real-life problems. Let’s explore some:

1️⃣ Identifying Even & Odd Numbers

Using b = 2 in Euclid’s lemma:

• If r = 0, the number is even (form: $2q$)
• If r = 1, the number is odd (form: $2q + 1$)

2️⃣ Identifying Forms of Odd Numbers

Using b = 4 in Euclid’s lemma:

• Odd numbers follow either $4q + 1$ or $4q + 3$

3️⃣ Real-World Example: Arranging Barfis 🍬

A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them such that:
✅ Each stack has the same number of barfis
✅ The total area occupied is minimum

👉 The best way to do this is by finding the HCF of 420 and 130.

Solution:

1️⃣ 420 ÷ 130 → Quotient = 3, Remainder = 30
👉 $420 = 130 \times 3 + 30$

2️⃣ 130 ÷ 30 → Quotient = 4, Remainder = 10
👉 $130 = 30 \times 4 + 10$

3️⃣ 30 ÷ 10 → Quotient = 3, Remainder = 0
👉 $30 = 10 \times 3 + 0$

🎯 The HCF is 10, meaning she should arrange the barfis in stacks of 10 each!

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🎯 Why is Euclid’s Algorithm Important?

✔️ Helps find HCF quickly (useful for large numbers)
✔️ Forms the base for advanced number theory
✔️ Used in computer algorithms for efficient calculations
✔️ Helps in real-life problems like grouping, tiling, and arranging objects

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🔹 Summary

• Euclid’s Division Lemma states that for any two numbers a and b, there exist unique numbers q (quotient) and r (remainder) such that: $$a = bq + r, \quad 0 \leq r < b$$
• Euclid’s Algorithm helps find the HCF of two numbers using repeated division.
• It has multiple applications in number theory, real-life grouping problems, and even computer science.

🔹 Remember: Euclid’s methods may be ancient, but they are still highly relevant today! 🚀✨

  Exercise 1.1.

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1. Finding the HCF Using Euclid’s Division Algorithm

Euclid’s Division Algorithm Statement:
For any two positive integers $a$ and $b$ (with $a \ge b$), there exist unique integers $q$ (the quotient) and $r$ (the remainder) such that

$$a = bq + r,\quad 0 \le r < b.$$

We repeat the process with $b$ and $r$ until $r=0$. The last nonzero remainder’s corresponding divisor is the HCF.

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(i) HCF of 135 and 225

1. Divide 225 by 135:

   $$225 = 135 \times 1 + 90.$$
   • Quotient: $q=1$
   • Remainder: $r=90$
     (Since $90 \neq 0$, we continue with 135 and 90.)

2. Divide 135 by 90:

   $$135 = 90 \times 1 + 45.$$
   • Quotient: $q=1$
   • Remainder: $r=45$
     (Since $45 \neq 0$, we continue with 90 and 45.)

3. Divide 90 by 45:

   $$90 = 45 \times 2 + 0.$$
   • Quotient: $q=2$
   • Remainder: $r=0$

Since the remainder is now 0, the divisor at this step (45) is the HCF.

$$\boxed{\text{HCF}(135, 225) = 45.}$$

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(ii) HCF of 196 and 38220

1. Divide 38220 by 196:

   $$38220 = 196 \times 195 + 0.$$
   • Quotient: $q=195$
   • Remainder: $r=0$

Since the remainder is 0 immediately, 196 is the HCF.

$$\boxed{\text{HCF}(196, 38220) = 196.}$$

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(iii) HCF of 867 and 255

1. Divide 867 by 255:

   $$867 = 255 \times 3 + 102.$$
   • Quotient: $q=3$
   • Remainder: $r=102$
     (Since $102 \neq 0$, continue with 255 and 102.)

2. Divide 255 by 102:

   $$255 = 102 \times 2 + 51.$$
   • Quotient: $q=2$
   • Remainder: $r=51$
     (Since $51 \neq 0$, continue with 102 and 51.)

3. Divide 102 by 51:

   $$102 = 51 \times 2 + 0.$$
   • Quotient: $q=2$
   • Remainder: $r=0$

Now the remainder is 0, so the divisor 51 is the HCF.

$$\boxed{\text{HCF}(867, 255) = 51.}$$

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2. Showing That Every Positive Odd Integer is of the Form $6q + 1$, $6q + 3$, or $6q + 5$

Any positive integer $a$ can be expressed using Euclid’s division lemma when divided by 6:

$$a = 6q + r,\quad \text{with } 0 \le r < 6.$$

The possible remainders $r$ are 0, 1, 2, 3, 4, or 5. Now, check the parity (even or odd) for each case:

• If $r = 0$: $a = 6q$ is even.
• If $r = 1$: $a = 6q + 1$ is odd.
• If $r = 2$: $a = 6q + 2$ is even.
• If $r = 3$: $a = 6q + 3$ is odd.
• If $r = 4$: $a = 6q + 4$ is even.
• If $r = 5$: $a = 6q + 5$ is odd.

Thus, for any odd integer, the remainder $r$ must be 1, 3, or 5, so every positive odd integer is in one of the following forms:

$$\boxed{6q + 1,\quad 6q + 3,\quad \text{or} \quad 6q + 5.}$$

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3. Maximum Number of Columns in a Parade

Problem Statement:
An army contingent of 616 members is to march behind an army band of 32 members. They must march in the same number of columns. To have an equal number of members in each column (with both groups arranged in the same number of columns), the number of columns must be a common divisor of both 616 and 32. The maximum such number is the HCF of 616 and 32.

Steps Using Euclid’s Algorithm:

1. Divide 616 by 32:

   $$616 = 32 \times 19 + 8.$$
   • Quotient: $q=19$
   • Remainder: $r=8$

2. Divide 32 by 8:

   $$32 = 8 \times 4 + 0.$$
   • Quotient: $q=4$
   • Remainder: $r=0$

Since the remainder is 0, the HCF is 8.

$$\boxed{\text{Maximum number of columns} = 8.}$$

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4. Showing That the Square of Any Positive Integer is Either of the Form $3m$ or $3m+1$

Let $x$ be any positive integer. By Euclid’s division lemma, when $x$ is divided by 3:

$$x = 3q + r,\quad \text{where } r \in \{0, 1, 2\}.$$

We now consider the square $x^2$ in each case.

Case 1: $r = 0$

If $x = 3q$, then

$$x^2 = (3q)^2 = 9q^2 = 3(3q^2).$$

Let $m = 3q^2$. Then

$$x^2 = 3m.$$

Case 2: $r = 1$

If $x = 3q + 1$, then

$$x^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1.$$

Let $m = 3q^2 + 2q$. Then

$$x^2 = 3m + 1.$$

Case 3: $r = 2$

If $x = 3q + 2$, then

$$x^2 = (3q+2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1.$$

Notice that

$$9q^2 + 12q + 3 = 3(3q^2 + 4q + 1).$$

Let $m = 3q^2 + 4q + 1$. Then

$$x^2 = 3m + 1.$$

Thus, in all cases, $x^2$ is either of the form $3m$ (when $r = 0$) or $3m+1$ (when $r = 1$ or $2$).

$$\boxed{x^2 = 3m \quad \text{or} \quad x^2 = 3m+1.}$$

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5. Showing That the Cube of Any Positive Integer is of the Form $9m$, $9m+1$, or $9m+8$

Again, let $x$ be any positive integer and write it as

$$x = 3q + r,\quad \text{where } r \in \{0, 1, 2\}.$$

We now consider $x^3$ for each value of $r$.

Case 1: $r = 0$

If $x = 3q$, then

$$x^3 = (3q)^3 = 27q^3 = 9(3q^3).$$

Let $m = 3q^3$. Then

$$x^3 = 9m.$$

Case 2: $r = 1$

If $x = 3q + 1$, then

$$x^3 = (3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1.$$

Let $m = 3q^3 + 3q^2 + q$. Then

$$x^3 = 9m + 1.$$

Case 3: $r = 2$

If $x = 3q + 2$, then

$$x^3 = (3q+2)^3 = 27q^3 + 54q^2 + 36q + 8.$$

Notice that

$$27q^3 + 54q^2 + 36q = 9(3q^3 + 6q^2 + 4q).$$

So we can write

$$x^3 = 9(3q^3 + 6q^2 + 4q) + 8.$$

Let $m = 3q^3 + 6q^2 + 4q$. Then

$$x^3 = 9m + 8.$$

Thus, for any integer $x$, the cube $x^3$ takes one of the forms:

$$\boxed{{\displaystyle x^3=9m,x^3=9m+1,\text{or}{x}^3=9m+8.\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}\mathrm{<br><span></span>}}}$$

📖 Detailed Notes on the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic (FTA) is one of the most important results in number theory. It states that every composite number can be uniquely expressed as a product of prime numbers, except for the order in which the primes appear.

🔹 What is the Fundamental Theorem of Arithmetic?

Theorem Statement:
🔸 Every composite number can be expressed (factorized) as a product of prime numbers, and this factorization is unique, except for the order of the factors.

This means that for any composite number, there is only one unique way to write it as a product of prime numbers (ignoring the order).

🔍 Understanding Through Examples:

Let’s take a few numbers and break them down into their prime factors:

• 2 → Already a prime number. ✅
• 4 → $4 = 2 \times 2 = 2^2$
• 253 → $253 = 11 \times 23$
• 32760 → $32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13$

No matter how you factorize 32760, you will always get the same prime numbers as factors, just in a different order.

🔢 Factor Tree Method for Prime Factorization

A factor tree is a visual way to break a number down into its prime factors.

Example: Factorizing 32760 using a Factor Tree

1. Start with 32760
2. Divide by 2 repeatedly: $$32760 \div 2 = 16380$$ $$16380 \div 2 = 8190$$ $$8190 \div 2 = 4095$$
3. Next, divide by 3: $$4095 \div 3 = 1365$$ $$1365 \div 3 = 455$$
4. Divide by 5: $$455 \div 5 = 91$$
5. Divide by 7: $$91 \div 7 = 13$$
6. 13 is prime, so we stop. ✅

$$32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13$$

Thus, 32760 is uniquely expressed as a product of primes.

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🔹 Why is FTA Important?

1️⃣ It ensures every composite number can be written as a product of primes.
2️⃣ It proves that this factorization is unique, meaning no matter how you factorize a number, the prime factors will always be the same.
3️⃣ It is the foundation for many other mathematical concepts, such as finding HCF (Highest Common Factor) and LCM (Least Common Multiple).

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🔹 Applications of the Fundamental Theorem of Arithmetic

1️⃣ Checking Whether a Number Can End in Zero

Example: Consider numbers of the form $4^n$. Can any of these end in 0?

Step 1: Prime Factorization of $4^n$

$$4^n = (2^2)^n = 2^{2n}$$

Since the only prime factor in $4^n$ is 2, it is not divisible by 5.

Step 2: Condition for a Number to End in 0
A number ends in 0 if it is divisible by 10, i.e., it has both 2 and 5 as factors.

Since $4^n$ does not contain 5, it can never end in 0.
Thus, no power of 4 can end in zero. ✅

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2️⃣ Finding HCF and LCM Using Prime Factorization

Example 1: Finding HCF and LCM of 6 and 20

Step 1: Prime Factorization

$$6 = 2^1 \times 3^1$$ $$20 = 2^2 \times 5^1$$

Step 2: Finding HCF
HCF is the product of the smallest power of each common prime factor:

$$HCF(6, 20) = 2^1 = 2.$$

Step 3: Finding LCM
LCM is the product of the greatest power of each prime factor:

$$LCM(6, 20) = 2^2 \times 3^1 \times 5^1 = 60.$$

Thus,

$$HCF(6, 20) = 2, \quad LCM(6, 20) = 60.$$

📝 Important Observation:

$$HCF(6,20) \times LCM(6,20) = 6 \times 20.$$

This shows that for any two numbers a and b:

$$HCF(a, b) \times LCM(a, b) = a \times b.$$

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3️⃣ Finding HCF and LCM of Larger Numbers

Example 2: HCF and LCM of 96 and 404

Step 1: Prime Factorization

$$96 = 2^5 \times 3$$ $$404 = 2^2 \times 101$$

Step 2: Find HCF

$$HCF(96, 404) = 2^2 = 4.$$

Step 3: Find LCM

$$LCM(96, 404) = \frac{96 \times 404}{HCF(96, 404)} = \frac{96 \times 404}{4} = 9696.$$

Thus,

$$HCF(96, 404) = 4, \quad LCM(96, 404) = 9696.$$

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4️⃣ Finding HCF and LCM of Three Numbers

Example 3: HCF and LCM of 6, 72, and 120

Step 1: Prime Factorization

$$6 = 2^1 \times 3^1$$ $$72 = 2^3 \times 3^2$$ $$120 = 2^3 \times 3^1 \times 5^1$$

Step 2: Find HCF

$$HCF(6, 72, 120) = 2^1 \times 3^1 = 6.$$

Step 3: Find LCM

$$LCM(6, 72, 120) = 2^3 \times 3^2 \times 5^1 = 360.$$

📝 Key Note:
For three numbers, the product of HCF and LCM is not equal to the product of the numbers:

$$HCF(6, 72, 120) \times LCM(6, 72, 120) \neq 6 \times 72 \times 120.$$

Unlike the case of two numbers, the HCF-LCM product relation does not hold for three or more numbers.

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🔹 Key Takeaways

📌 Every composite number can be expressed uniquely as a product of prime numbers.
📌 The Fundamental Theorem of Arithmetic is crucial for finding HCF and LCM using prime factorization.
📌 The relation $HCF(a, b) \times LCM(a, b) = a \times b$ holds only for two numbers, not for three or more.
📌 Numbers that don’t have 5 in their prime factorization (e.g., $4^n$) can never end in 0.

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💡 Why is the Fundamental Theorem of Arithmetic Important?

✅ Used in Cryptography (like RSA encryption).
✅ Helps in simplifying fractions and finding greatest common divisors (GCD).
✅ Used in computer algorithms for prime factorization and efficient calculations.
✅ Foundation for many theorems in number theory and modern mathematics.

💡 Mastering this concept helps in solving advanced mathematical problems! 🚀

📖 Exercise 1.2 – Questions & Answers

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1️⃣ Express Each Number as a Product of Its Prime Factors

(i) 140

Answer:

$$140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$$

(ii) 156

Answer:

$$156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$$

(iii) 3825

Answer:

$$3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$$

(iv) 5005

Answer:

$$5005 = 5 \times 7 \times 11 \times 13$$

(v) 7429

Answer:

$$7429 = 17 \times 19 \times 23$$

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2️⃣ Find the LCM and HCF of the Following Pairs of Integers and Verify That

$$\text{HCF} \times \text{LCM} = \text{Product of the Two Numbers}$$

(i) 26 and 91

Step 1: Prime Factorization

$$26 = 2 \times 13, \quad 91 = 7 \times 13$$

Step 2: Find HCF

$$HCF(26, 91) = 13$$

Step 3: Find LCM

$$LCM(26, 91) = 2 \times 7 \times 13 = 182$$

Step 4: Verify

$$HCF \times LCM = 13 \times 182 = 2366$$ $$26 \times 91 = 2366$$

✅ Verified!

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(ii) 510 and 92

Step 1: Prime Factorization

$$510 = 2 \times 3 \times 5 \times 17, \quad 92 = 2^2 \times 23$$

Step 2: Find HCF

$$HCF(510, 92) = 2$$

Step 3: Find LCM

$$LCM(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$$

Step 4: Verify

$$HCF \times LCM = 2 \times 23460 = 46920$$ $$510 \times 92 = 46920$$

✅ Verified!

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(iii) 336 and 54

Step 1: Prime Factorization

$$336 = 2^4 \times 3 \times 7, \quad 54 = 2 \times 3^3$$

Step 2: Find HCF

$$HCF(336, 54) = 2 \times 3 = 6$$

Step 3: Find LCM

$$LCM(336, 54) = 2^4 \times 3^3 \times 7 = 3024$$

Step 4: Verify

$$HCF \times LCM = 6 \times 3024 = 18144$$ $$336 \times 54 = 18144$$

✅ Verified!

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3️⃣ Find the LCM and HCF of the Following Integers by Prime Factorization

(i) 12, 15, and 21

$$12 = 2^2 \times 3, \quad 15 = 3 \times 5, \quad 21 = 3 \times 7$$

HCF:

$$HCF(12, 15, 21) = 3$$

LCM:

$$LCM(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$$

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(ii) 17, 23, and 29

Since 17, 23, and 29 are all prime numbers,

$$HCF(17, 23, 29) = 1$$ $$LCM(17, 23, 29) = 17 \times 23 \times 29 = 11339$$

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(iii) 8, 9, and 25

$$8 = 2^3, \quad 9 = 3^2, \quad 25 = 5^2$$

HCF:

$$HCF(8, 9, 25) = 1$$

LCM:

$$LCM(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 1800$$

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4️⃣ Given That HCF(306, 657) = 9, Find LCM(306, 657)

Using the formula:

$$HCF(a, b) \times LCM(a, b) = a \times b$$ $$9 \times LCM(306, 657) = 306 \times 657$$ $$LCM(306, 657) = \frac{306 \times 657}{9} = 22338$$

✅ LCM(306, 657) = 22338.

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5️⃣ Check Whether $6^n$ Can End with 0 for Any Natural Number $n$

For a number to end in 0, it must be divisible by 10 (i.e., have both 2 and 5 as prime factors).

Since

$$6^n = (2 \times 3)^n = 2^n \times 3^n,$$

it does not contain 5, so it can never end in 0.

✅ Thus, $6^n$ can never end in 0.

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6️⃣ Explain Why These Numbers Are Composite

(i) $7 \times 11 \times 13 + 13$

Factor out 13:

$$7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1) = 13 \times 78.$$

Since this number has factors other than 1 and itself, it is composite. ✅

(ii) $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

Factor out 5:

$$5 (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1).$$

Since this number has factors other than 1 and itself, it is composite. ✅

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7️⃣ After How Many Minutes Will Sonia and Ravi Meet Again?

• Sonia takes 18 minutes per round.
• Ravi takes 12 minutes per round.
• They will meet again after the LCM of 18 and 12.

Finding LCM

$$18 = 2 \times 3^2, \quad 12 = 2^2 \times 3$$ $$LCM(18, 12) = 2^2 \times 3^2 = 36.$$

✅ Sonia and Ravi will meet again after 36 minutes. ✅

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🔹 Final Summary

✔ HCF = highest common factor, LCM = smallest common multiple.
✔ HCF × LCM = Product of two numbers (for two numbers only).
✔ A number ending in 0 must have 5 as a factor.
✔ Composite numbers have more than two factors.
✔ LCM helps solve real-world problems like meeting times.

💡 Understanding these concepts makes problem-solving easy! 🚀

📖 Detailed Notes on Revisiting Irrational Numbers (Exercise 1.4)

🔹 Introduction to Irrational Numbers

• In Class IX, we learned about irrational numbers and how they fit into the real number system.
• A number is irrational if it cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
• Examples of irrational numbers:
  • $\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi, e$
  • Non-repeating, non-terminating decimals like 0.10110111011110...

Now, we will prove that numbers like $\sqrt{2}, \sqrt{3}, \sqrt{5}$, and any prime number $p$ in square root form are irrational.

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🔹 Theorem 1.3: If a Prime Number $p$ Divides $a^2$, Then $p$ Also Divides $a$

Statement: If $p$ is a prime number and $p$ divides $a^2$, then $p$ must also divide $a$.

✅ Proof Using the Fundamental Theorem of Arithmetic

1️⃣ Any number $a$ can be expressed in prime factorized form as:

$$a = p_1 p_2 \dots p_n$$

where $p_1, p_2, \dots, p_n$ are prime numbers.

2️⃣ Squaring both sides:

$$a^2 = (p_1 p_2 \dots p_n) \times (p_1 p_2 \dots p_n) = p_1^2 p_2^2 \dots p_n^2$$

3️⃣ We are given that $p$ divides $a^2$.

4️⃣ From the Fundamental Theorem of Arithmetic, $p$ must be one of the prime factors of $a^2$.

5️⃣ Since the same prime factors appear in $a$, $p$ must divide $a$.

✅ Thus, if a prime number divides the square of a number, it must also divide the number itself.

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🔹 Theorem 1.4: Proof That $\sqrt{2}$ is Irrational

We use proof by contradiction:

✅ Proof

1️⃣ Assume that $\sqrt{2}$ is rational.

• This means we can write it as a fraction:

$$\sqrt{2} = \frac{r}{s}$$

where $r$ and $s$ are coprime integers (i.e., they have no common factors other than 1).

2️⃣ Squaring both sides:

$$2 = \frac{r^2}{s^2}$$

Multiplying by $s^2$:

$$2s^2 = r^2$$

So, $r^2$ is divisible by 2.

3️⃣ Using Theorem 1.3, since $2$ divides $r^2$, it must also divide $r$.

• Let $r = 2c$ for some integer $c$.

4️⃣ Substituting $r = 2c$ into $2s^2 = r^2$:

$$2s^2 = (2c)^2 = 4c^2$$

Dividing by 2:

$$s^2 = 2c^2$$

This means $s^2$ is also divisible by 2, so $s$ is divisible by 2 (by Theorem 1.3).

5️⃣ Contradiction!

• We assumed that $r$ and $s$ were coprime, but both are divisible by 2.
• This contradicts our assumption.

✅ Conclusion: $\sqrt{2}$ must be irrational.

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🔹 Example: Proof That $\sqrt{3}$ is Irrational

We use the same contradiction method as before.

✅ Proof

1️⃣ Assume $\sqrt{3}$ is rational, so we can write:

$$\sqrt{3} = \frac{a}{b}$$

where $a$ and $b$ are coprime.

2️⃣ Squaring both sides:

$$3b^2 = a^2$$

So, $a^2$ is divisible by 3.

3️⃣ Using Theorem 1.3, $a$ is also divisible by 3.

• Let $a = 3c$ for some integer $c$.

4️⃣ Substituting $a = 3c$ into $3b^2 = a^2$:

$$3b^2 = (3c)^2 = 9c^2$$

Dividing by 3:

$$b^2 = 3c^2$$

So, $b^2$ is divisible by 3, meaning $b$ is also divisible by 3.

5️⃣ Contradiction!

• We assumed $a$ and $b$ were coprime, but both are divisible by 3.
• This contradicts our assumption.

✅ Conclusion: $\sqrt{3}$ must be irrational.

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🔹 Important Properties of Irrational Numbers

1️⃣ Sum/Difference of a Rational and an Irrational Number is Irrational

• Example: $5 - \sqrt{3}$ is irrational.

✅ Proof:

• Assume $5 - \sqrt{3}$ is rational, say $\frac{a}{b}$.
• Rearranging: $$\sqrt{3} = 5 - \frac{a}{b}$$
• Since $5 - \frac{a}{b}$ is rational, $\sqrt{3}$ must be rational.
• But $\sqrt{3}$ is irrational (contradiction!).

✅ Conclusion: $5 - \sqrt{3}$ is irrational.

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2️⃣ Product/Quotient of a Non-Zero Rational and an Irrational is Irrational

• Example: $3\sqrt{2}$ is irrational.

✅ Proof:

• Assume $3\sqrt{2}$ is rational, say $\frac{a}{b}$.
• Rearranging: $$\sqrt{2} = \frac{a}{3b}$$
• Since $\frac{a}{3b}$ is rational, $\sqrt{2}$ must be rational (contradiction!).

✅ Conclusion: $3\sqrt{2}$ is irrational.

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🔹 Summary of Key Points

✅ Definition of Irrational Numbers: Cannot be written as $\frac{p}{q}$, where $p$ and $q$ are integers.

✅ Theorem 1.3: If a prime $p$ divides $a^2$, it must divide $a$.

✅ Proof by Contradiction for $\sqrt{2}$ and $\sqrt{3}$ Being Irrational.

✅ Sum/Difference of a Rational and an Irrational is Irrational.

✅ Product/Quotient of a Rational (non-zero) and an Irrational is Irrational.

💡 Understanding these proofs strengthens your knowledge of number properties! 🚀

📖 Exercise 1.3 – Questions & Answers

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1️⃣ Prove that $\sqrt{5}$ is Irrational

We use proof by contradiction.

✅ Proof

1️⃣ Assume $\sqrt{5}$ is rational.

• This means it can be written as: $$\sqrt{5} = \frac{a}{b}$$ where $a$ and $b$ are coprime integers (i.e., they have no common factors other than 1).

2️⃣ Squaring both sides:

$$5 = \frac{a^2}{b^2}$$

Multiplying by $b^2$:

$$5b^2 = a^2$$

So, $a^2$ is divisible by 5.

3️⃣ Using Theorem 1.3, since $5$ divides $a^2$, it must also divide $a$.

• Let $a = 5c$ for some integer $c$.

4️⃣ Substituting $a = 5c$ into $5b^2 = a^2$:

$$5b^2 = (5c)^2 = 25c^2$$

Dividing by 5:

$$b^2 = 5c^2$$

This means $b^2$ is divisible by 5, so $b$ is divisible by 5 (by Theorem 1.3).

5️⃣ Contradiction!

• We assumed that $a$ and $b$ were coprime, but both are divisible by 5.
• This contradicts our assumption.

✅ Conclusion: $\sqrt{5}$ must be irrational.

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2️⃣ Prove that $3 + 2\sqrt{5}$ is Irrational

We use proof by contradiction.

✅ Proof

1️⃣ Assume that $3 + 2\sqrt{5}$ is rational.

• That is, we can write: $$3 + 2\sqrt{5} = \frac{a}{b}$$ where $a, b$ are integers, and $b \neq 0$.

2️⃣ Rearrange the equation:

$$2\sqrt{5} = \frac{a}{b} - 3$$ $$\sqrt{5} = \frac{a - 3b}{2b}$$

Since $a, b$ are integers, $\frac{a - 3b}{2b}$ is rational.

3️⃣ But this contradicts the fact that $\sqrt{5}$ is irrational.

✅ Conclusion: $3 + 2\sqrt{5}$ must be irrational.

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3️⃣ Prove That the Following Are Irrational

(i) $\frac{1}{\sqrt{2}}$ is Irrational

1️⃣ Assume $\frac{1}{\sqrt{2}}$ is rational.

• That means: $$\frac{1}{\sqrt{2}} = \frac{a}{b}$$ where $a, b$ are integers, and $b \neq 0$.

2️⃣ Multiply both sides by $\sqrt{2}$:

$$1 = \frac{a \sqrt{2}}{b}$$ $$\sqrt{2} = \frac{b}{a}$$

Since $\frac{b}{a}$ is rational, this contradicts the fact that $\sqrt{2}$ is irrational.

✅ Conclusion: $\frac{1}{\sqrt{2}}$ is irrational.

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(ii) $7\sqrt{5}$ is Irrational

1️⃣ Assume $7\sqrt{5}$ is rational.

• That means: $$7\sqrt{5} = \frac{a}{b}$$ where $a, b$ are integers, and $b \neq 0$.

2️⃣ Rearrange the equation:

$$\sqrt{5} = \frac{a}{7b}$$

Since $\frac{a}{7b}$ is rational, this contradicts the fact that $\sqrt{5}$ is irrational.

✅ Conclusion: $7\sqrt{5}$ is irrational.

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(iii) $6 + 2\sqrt{3}$ is Irrational

1️⃣ Assume $6 + 2\sqrt{3}$ is rational.

• That means: $$6 + 2\sqrt{3} = \frac{a}{b}$$ where $a, b$ are integers, and $b \neq 0$.

2️⃣ Rearrange the equation:

$$2\sqrt{3} = \frac{a}{b} - 6$$ $$\sqrt{3} = \frac{a - 6b}{2b}$$

Since $\frac{a - 6b}{2b}$ is rational, this contradicts the fact that $\sqrt{3}$ is irrational.

✅ Conclusion: $6 + 2\sqrt{3}$ is irrational.

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🔹 Summary of Key Concepts

✅ Definition of Irrational Numbers: Cannot be written as $\frac{p}{q}$, where $p$ and $q$ are integers.

✅ Proof by Contradiction Method: Assume the number is rational and show that it leads to a contradiction.

✅ Key Results:

• $\sqrt{5}$ is irrational.
• $3 + 2\sqrt{5}$ is irrational.
• $\frac{1}{\sqrt{2}}, 7\sqrt{5}, 6 + 2\sqrt{3}$ are all irrational.

💡 Understanding these proofs strengthens your knowledge of number properties! 🚀

📖 Easy Notes on Rational Numbers and Their Decimal Expansions (Exercise 1.5)

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🔹 What Are Rational Numbers?

• Rational numbers are numbers that can be written in the form $\frac{p}{q}$, where:

  • $p$ and $q$ are integers (whole numbers).
  • $q \neq 0$ (denominator cannot be zero).

• Rational numbers have two types of decimal expansions:
  1️⃣ Terminating Decimal → The decimal stops after some digits.
  2️⃣ Non-Terminating, Repeating Decimal → The decimal never ends but follows a repeating pattern.

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🔹 What is a Terminating Decimal Expansion?

A terminating decimal has a fixed number of decimal places and then stops.

✅ Examples

1️⃣ $0.375$ → Stops after three decimal places.
2️⃣ $0.104$ → Stops after three decimal places.
3️⃣ $23.3408$ → Stops after four decimal places.

These numbers can be written as fractions:

$$0.375 = \frac{375}{1000} = \frac{3}{8}$$ $$0.104 = \frac{104}{1000} = \frac{13}{125}$$ $$23.3408 = \frac{233408}{10000}$$

After simplifying, we notice that the denominator of each fraction only contains 2 and 5.

✅ Rule: A rational number has a terminating decimal expansion if the denominator contains only the prime numbers 2 and 5.

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🔹 Theorem 1.5: Rule for Terminating Decimals

If a rational number $\frac{p}{q}$ has a terminating decimal, then the denominator $q$ can be written as:

$$q = 2^n \times 5^m$$

where $n$ and $m$ are non-negative whole numbers (0, 1, 2, 3, ...).

✅ Example:

$$\frac{7}{40} = \frac{7}{2^3 \times 5}$$

Since the denominator only contains 2 and 5, the decimal expansion terminates.

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🔹 What is a Non-Terminating, Repeating Decimal?

A non-terminating, repeating decimal never stops but follows a pattern that repeats.

✅ Example

$$\frac{1}{7} = 0.142857142857...$$

• The decimal keeps going but repeats the pattern 142857.
• The denominator (7) does not have only 2 and 5 as factors.

✅ Rule: A rational number has a non-terminating, repeating decimal expansion if the denominator contains prime factors other than 2 and 5.

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🔹 Theorem 1.6: Another Rule for Terminating Decimals

If a rational number $\frac{p}{q}$ has a denominator in the form $2^n \times 5^m$, then it will always have a terminating decimal expansion.

✅ Example:

$$\frac{13}{125} = 0.104$$

Since the denominator is $5^3$ (only 5s), the decimal expansion terminates.

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🔹 Theorem 1.7: Rule for Non-Terminating, Repeating Decimals

If a rational number $\frac{p}{q}$ has a denominator that contains other prime numbers besides 2 and 5, then the decimal expansion is non-terminating, repeating.

✅ Example:

$$\frac{1}{7} = 0.142857142857...$$

Since 7 is not 2 or 5, the decimal does not terminate and keeps repeating.

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🔹 Summary of Key Points

✔ Terminating Decimal Expansion: If the denominator only contains 2 and 5, the decimal stops.
✔ Non-Terminating, Repeating Decimal Expansion: If the denominator contains other prime numbers, the decimal keeps repeating.
✔ Key Theorems:

• Theorem 1.5: If the decimal expansion stops, the denominator is $2^n \times 5^m$.
• Theorem 1.6: If the denominator is $2^n \times 5^m$, the decimal stops.
• Theorem 1.7: If the denominator has other numbers, the decimal keeps repeating.

💡 Conclusion: Every rational number has either a terminating or a repeating decimal expansion. 🚀

📖 Exercise 1.4 – Questions & Answers

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1️⃣ Without performing the long division, state whether the following rational numbers will have a terminating or non-terminating repeating decimal expansion.

Rule to Remember:

A rational number $\frac{p}{q}$ has:

• A terminating decimal expansion if $q$ (denominator) has only 2 and/or 5 as prime factors.
• A non-terminating repeating decimal expansion if $q$ has any prime factors other than 2 and 5.

(i) $\frac{13}{3125}$

• Prime factorization of 3125: $$3125 = 5^5$$
• Contains only 5s → ✅ Terminating decimal expansion

(ii) $\frac{17}{8}$

• Prime factorization of 8: $$8 = 2^3$$
• Contains only 2s → ✅ Terminating decimal expansion

(iii) $\frac{64}{455}$

• Prime factorization of 455: $$455 = 5 \times 7 \times 13$$
• Contains 7 and 13 (other than 2 and 5) → ❌ Non-terminating, repeating decimal expansion

(iv) $\frac{15}{1600}$

• Prime factorization of 1600: $$1600 = 2^6 \times 5^2$$
• Contains only 2s and 5s → ✅ Terminating decimal expansion

(v) $\frac{29}{343}$

• Prime factorization of 343: $$343 = 7^3$$
• Contains 7 (not just 2s and 5s) → ❌ Non-terminating, repeating decimal expansion

(vi) $\frac{2^3}{2^3 \times 5}$

• Simplifies to $\frac{8}{40} = \frac{1}{5}$
• Denominator = 5 → ✅ Terminating decimal expansion

(vii) $\frac{2^7 \times 5}{2^9 \times 5 \times 7}$

• Denominator = $2^9 \times 5 \times 7$
• Contains 7 (besides 2 and 5) → ❌ Non-terminating, repeating decimal expansion

(viii) $\frac{6}{15}$

• Prime factorization of 15: $$15 = 3 \times 5$$
• Contains 3 (besides 2 and 5) → ❌ Non-terminating, repeating decimal expansion

(ix) $\frac{35}{50}$

• Prime factorization of 50: $$50 = 2 \times 5^2$$
• Contains only 2s and 5s → ✅ Terminating decimal expansion

(x) $\frac{77}{210}$

• Prime factorization of 210: $$210 = 2 \times 3 \times 5 \times 7$$
• Contains 3 and 7 (besides 2 and 5) → ❌ Non-terminating, repeating decimal expansion

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2️⃣ Write the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) $\frac{13}{3125}$

$$\frac{13}{3125} = 0.00416$$

(ii) $\frac{17}{8}$

$$\frac{17}{8} = 2.125$$

(iv) $\frac{15}{1600}$

$$\frac{15}{1600} = 0.009375$$

(vi) $\frac{8}{40} = \frac{1}{5}$

$$\frac{1}{5} = 0.2$$

(ix) $\frac{35}{50}$

$$\frac{35}{50} = 0.7$$

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3️⃣ The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form $\frac{p}{q}$, what can you say about the prime factors of $q$?

(i) 43.123456789

• The decimal stops after 9 places → ✅ Rational
• Can be written as a fraction: $$43.123456789 = \frac{43123456789}{1000000000}$$
• Denominator = $10^9 = 2^9 \times 5^9$
• Only contains 2s and 5s → ✅ Terminating decimal

✅ Conclusion: This is a rational number.

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(ii) 0.120120012000120000...

• The decimal does not have a repeating pattern.
• It keeps changing and does not repeat exactly.

❌ Conclusion: This is an irrational number.

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(iii) 43.123456789 (same as first case)

• The decimal stops after 9 places → ✅ Rational
• Denominator contains only 2s and 5s → Terminating decimal.

✅ Conclusion: This is a rational number.

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🔹 Final Summary

✔ A rational number has either a terminating or a repeating decimal expansion.
✔ If the denominator of $\frac{p}{q}$ contains only 2s and/or 5s, it has a terminating decimal expansion.
✔ If the denominator contains other prime numbers, it has a non-terminating, repeating decimal expansion.
✔ Non-repeating, non-terminating decimals are irrational numbers.

💡 Understanding these rules helps in quickly identifying rational and irrational numbers! 🚀

It looks like you're referring to an important mathematical property related to the HCF (Highest Common Factor) and LCM (Least Common Multiple) of three numbers. Below is a clear and structured explanation of the given results.

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📖 Understanding HCF and LCM for Three Numbers

Key Concept

For two numbers $p$ and $q$, we know that:

$$HCF(p, q) \times LCM(p, q) = p \times q$$

However, for three numbers $p, q, r$, this formula does not hold:

$$HCF(p, q, r) \times LCM(p, q, r) \neq p \times q \times r$$

Instead, we use two special formulas for finding HCF and LCM of three numbers.

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📌 Formula for LCM of Three Numbers

The LCM (Least Common Multiple) of three numbers is given by:

$$LCM (p, q, r) = \frac{HCF(p, q) \times HCF(q, r) \times HCF(p, r) \times p \times q \times r}{HCF(p, q, r)}$$

✅ What This Means:
This formula ensures that we correctly find the smallest number that is divisible by $p, q,$ and $r$ while considering their common factors.

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📌 Formula for HCF of Three Numbers

The HCF (Highest Common Factor) of three numbers is given by:

$$HCF (p, q, r) = \frac{LCM(p, q) \times LCM(q, r) \times LCM(p, r)}{LCM(p, q, r)}$$

✅ What This Means:
This formula ensures that we correctly find the largest number that divides all three numbers.

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📌 Example Calculation

Let's apply these formulas to three numbers: 6, 9, and 12.

1️⃣ Step 1: Find HCF of pairs

• $HCF(6,9) = 3$
• $HCF(9,12) = 3$
• $HCF(6,12) = 6$
• $HCF(6,9,12) = 3$

2️⃣ Step 2: Find LCM of pairs

• $LCM(6,9) = 18$
• $LCM(9,12) = 36$
• $LCM(6,12) = 12$
• $LCM(6,9,12) = 36$

3️⃣ Step 3: Verify formulas

• Using the LCM formula: $$LCM(6,9,12) = \frac{3 \times 3 \times 6 \times 6 \times 9 \times 12}{3} = 36$$
• Using the HCF formula: $$HCF(6,9,12) = \frac{18 \times 36 \times 12}{36} = 3$$

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🔹 Summary

✔ HCF and LCM for three numbers do not follow the same simple rule as for two numbers.
✔ The special formulas account for how numbers share factors across three values.
✔ These formulas are useful in problems involving LCM and HCF in real-life applications like time cycles, gear rotations, and resource management.

💡 Understanding these relationships helps in solving advanced number system problems efficiently!