## Chapter -1 (Real Number)

Real numbers are a fundamental concept in mathematics that encompasses all the numbers that can be represented on the number line. They include both rational numbers (fractions) and irrational numbers (numbers that cannot be expressed as fractions) (Euclid’s division).

Real numbers can be positive, negative, or zero, and they can also be represented as decimals or fractions. Examples of real numbers include 3, -2, 0, 1.5, √2 (the square root of 2), and π (pi).

The set of real numbers is vast and continuous, with an infinite number of points between any two real numbers. This property is known as the density of real numbers. It means that no matter how close two real numbers are, there will always be another real number between them.

Real numbers are used extensively in various branches of mathematics, including algebra, calculus, geometry, and analysis. They form the foundation for mathematical operations and serve as a basis for solving equations, modeling real-world phenomena, and studying the properties of functions and geometric figures.

In summary, real numbers encompass all the rational and irrational numbers, providing a comprehensive representation of quantities and measurements along the number line. They are a fundamental concept in mathematics and play a crucial role in numerous mathematical applications and disciplines.

## HCF Finder

**NCERT - Solution of Class 10 Mathematics Chapter -1**

### Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of:

**i. 135 and 225 ii. 196 and 38220 iii. 867 and 255**

**Solutions**:

i. Let's find the HCF (Highest Common Factor) of 135 and 225 using Euclid's division algorithm.

Step 1: Divide 225 by 135. 225 = 135 * 1 + 90

Step 2: Now, divide 135 by 90. 135 = 90 * 1 + 45

Step 3: Divide 90 by 45. 90 = 45 * 2 + 0

Since the remainder has become zero, we stop. The divisor at this point, which is 45, is the HCF of 135 and 225.

Therefore, the HCF of 135 and 225 is 45.

ii. Let's find the HCF of 196 and 38220 using Euclid's division algorithm.

Step 1: Divide 38220 by 196. 38220 = 196 * 195 + 0

Since the remainder is zero, we stop. The divisor at this point, which is 196, is the HCF of 196 and 38220.

Therefore, the HCF of 196 and 38220 is 196.

iii. Let's find the HCF of 867 and 255 using Euclid's division algorithm.

Step 1: Divide 867 by 255. 867 = 255 * 3 + 102

Step 2: Now, divide 255 by 102. 255 = 102 * 2 + 51

Step 3: Divide 102 by 51. 102 = 51 * 2 + 0

Since the remainder has become zero, we stop. The divisor at this point, which is 51, is the HCF of 867 and 255.

Therefore, the HCF of 867 and 255 is 51.

**2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.**

**Solution:**

To prove that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer, we can consider two cases:

Case 1: The odd integer is of the form 6q + 1.

Let's assume an odd integer x = 6q + 1, where q is an integer. We can express x as:

x = (6q + 1) = 6q + (6q + 1) = 6q + 6q + 1 = 12q + 1 = 6(2q) + 1

Therefore, any odd integer of the form 6q + 1 can be written in the given form.

Case 2: The odd integer is not of the form 6q + 1.

Let's assume an odd integer y that is not of the form 6q + 1. In this case, it can be expressed as:

y = 6q + r

Here, r represents the remainder when dividing y by 6. Since y is an odd integer, the possible remainders when dividing by 6 are 1, 3, or 5. Therefore, the odd integer y can be expressed as 6q + 1, 6q + 3, or 6q + 5, depending on the remainder.

Hence, any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.

**3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?**

**Solution:**

To find the maximum number of columns in which the army contingent and the army band can march, we need to determine the greatest common divisor (GCD) of the two group sizes: 616 and 32.

Using Euclid's algorithm, we can find the GCD:

Step 1: Divide 616 by 32.

616 = 32 * 19 + 8

Step 2: Divide 32 by 8.

32 = 8 * 4 + 0

Since the remainder has become zero, we stop. The divisor at this point, which is 8, is the GCD of 616 and 32.

Therefore, the maximum number of columns in which the army contingent and the army band can march is 8.

**4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.**

**Solution:**

To show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m, we can use Euclid's division lemma.

According to Euclid's division lemma, for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Let's consider the square of a positive integer:

(a)^2 = a * a

Case 1: If a is divisible by 3, i.e., a = 3k for some integer k:

In this case, (a)^2 = (3k)^2 = 9k^2 = 3(3k^2), which is of the form 3m, where m = 3k^2 is an integer.

Case 2: If a is not divisible by 3, i.e., a = 3k + 1 or a = 3k + 2 for some integer k:

In this case, (a)^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1, which is of the form 3m + 1, where m = 3k^2 + 2k is an integer.

Therefore, for any positive integer a, the square of a is either of the form 3m or 3m + 1 for some integer m.

**5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.**

**Solution:**

To show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8 for some integer m, we can utilize Euclid's division lemma.

According to Euclid's division lemma, for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Let's consider the cube of a positive integer:

(a)^3 = a * a * a

Case 1: If a is divisible by 3, i.e., a = 3k for some integer k:

In this case, (a)^3 = (3k)^3 = 27k^3 = 9(3k^3) = 9m, where m = 3k^3 is an integer.

Case 2: If a leaves a remainder of 1 when divided by 3, i.e., a = 3k + 1 for some integer k:

In this case, (a)^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 = 9(3k^3 + 3k^2 + k) + 1, which is of the form 9m + 1, where m = 3k^3 + 3k^2 + k is an integer.

Case 3: If a leaves a remainder of 2 when divided by 3, i.e., a = 3k + 2 for some integer k:

In this case, (a)^3 = (3k + 2)^3 = 27k^3 + 54k^2 + 36k + 8 = 9(3k^3 + 6k^2 + 4k) + 8, which is of the form 9m + 8, where m = 3k^3 + 6k^2 + 4k is an integer.

Therefore, for any positive integer a, the cube of a is of the form 9m, 9m + 1, or 9m + 8 for some integer m.

**Exercise 1.2**

**1. Express each number as a product of its prime factors:**

**(i) 140**

**(ii) 156**

**(iii) 3825**

**(iv) 5005**

**(v) 7429**

**Solution:**

(i) To express 140 as a product of its prime factors, we can perform prime factorization:

140 = 2 * 70

= 2 * 2 * 35

= 2 * 2 * 5 * 7

Therefore, the prime factorization of 140 is 2^2 * 5 * 7.

(ii) To express 156 as a product of its prime factors:

156 = 2 * 78

= 2 * 2 * 39

= 2 * 2 * 3 * 13

Therefore, the prime factorization of 156 is 2^2 * 3 * 13.

(iii) To express 3825 as a product of its prime factors:

3825 = 3 * 1275

= 3 * 3 * 425

= 3 * 3 * 5 * 85

= 3 * 3 * 5 * 5 * 17

Therefore, the prime factorization of 3825 is 3^2 * 5^2 * 17.

(iv) To express 5005 as a product of its prime factors:

5005 = 5 * 1001

= 5 * 7 * 143

= 5 * 7 * 11 * 13

Therefore, the prime factorization of 5005 is 5 * 7 * 11 * 13.

(v) To express 7429 as a product of its prime factors:

7429 is a prime number, so its prime factorization is simply 7429.

Therefore, the prime factorization of 7429 is 7429 itself.

**2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.**

**(i) 26 and 91**

**(ii) 510 and 92**

**(iii) 336 and 54**

**Solution:**

(i) To find the LCM and HCF of 26 and 91:

Prime factorization of 26: 2 * 13

Prime factorization of 91: 7 * 13

HCF of 26 and 91: The highest common factor (HCF) is 13.

LCM of 26 and 91: The least common multiple (LCM) can be found by multiplying the highest power of each prime factor that appears in either number. In this case, the LCM is 2 * 7 * 13 = 182.

We can verify that LCM × HCF = product of the two numbers:

182 × 13 = 2366, which is equal to 26 × 91.

Therefore, the LCM and HCF of 26 and 91 are 182 and 13, respectively.

(ii) To find the LCM and HCF of 510 and 92:

Prime factorization of 510: 2 * 3 * 5 * 17

Prime factorization of 92: 2 * 2 * 23

HCF of 510 and 92: The highest common factor (HCF) is 2.

LCM of 510 and 92: The least common multiple (LCM) is 2 * 2 * 3 * 5 * 17 * 23 = 19,380.

We can verify that LCM × HCF = product of the two numbers:

19,380 × 2 = 38,760, which is equal to 510 × 92.

Therefore, the LCM and HCF of 510 and 92 are 19,380 and 2, respectively.

(iii) To find the LCM and HCF of 336 and 54:

Prime factorization of 336: 2 * 2 * 2 * 2 * 3 * 7

Prime factorization of 54: 2 * 3 * 3 * 3

HCF of 336 and 54: The highest common factor (HCF) is 6.

LCM of 336 and 54: The least common multiple (LCM) is 2 * 2 * 2 * 2 * 3 * 3 * 3 * 7 = 1,512.

We can verify that LCM × HCF = product of the two numbers:

1,512 × 6 = 9,072, which is equal to 336 × 54.

Therefore, the LCM and HCF of 336 and 54 are 1,512 and 6, respectively.

**3. Find the LCM and HCF of the following integers by applying the prime factorisation method.**

**(i) 12, 15 and 21**

**(ii) 17, 23 and 29**

**(iii) 8, 9 and 25**

**Solutions:**

(i) To find the LCM and HCF of 12, 15, and 21:

Prime factorization of 12: 2 * 2 * 3

Prime factorization of 15: 3 * 5

Prime factorization of 21: 3 * 7

HCF of 12, 15, and 21: The highest common factor (HCF) is 3.

LCM of 12, 15, and 21: The least common multiple (LCM) can be found by multiplying the highest power of each prime factor that appears in any of the numbers. In this case, the LCM is 2 * 2 * 3 * 5 * 7 = 420.

Therefore, the LCM and HCF of 12, 15, and 21 are 420 and 3, respectively.

(ii) To find the LCM and HCF of 17, 23, and 29:

Since 17, 23, and 29 are prime numbers, their prime factorizations are themselves.

HCF of 17, 23, and 29: The highest common factor (HCF) is 1 since they have no common factors.

LCM of 17, 23, and 29: The least common multiple (LCM) is the product of the three numbers, which is 17 * 23 * 29 = 11,879.

Therefore, the LCM and HCF of 17, 23, and 29 are 11,879 and 1, respectively.

(iii) To find the LCM and HCF of 8, 9, and 25:

Prime factorization of 8: 2 * 2 * 2

Prime factorization of 9: 3 * 3

Prime factorization of 25: 5 * 5

HCF of 8, 9, and 25: The highest common factor (HCF) is 1 since they have no common factors.

LCM of 8, 9, and 25: The least common multiple (LCM) can be found by multiplying the highest power of each prime factor that appears in any of the numbers. In this case, the LCM is 2 * 2 * 2 * 3 * 3 * 5 * 5 = 900.

Therefore, the LCM and HCF of 8, 9, and 25 are 900 and 1, respectively.

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

**Solution:**

To find the LCM (Least Common Multiple) of 306 and 657 when their HCF (Highest Common Factor) is known to be 9, we can use the formula:

LCM (a, b) = (a * b) / HCF (a, b)

Given HCF (306, 657) = 9, we can substitute the values into the formula:

LCM (306, 657) = (306 * 657) / 9

Now we can calculate the LCM:

LCM (306, 657) = (201,342) / 9

LCM (306, 657) = 22,371

Therefore, the LCM of 306 and 657, given HCF (306, 657) = 9, is 22,371.

**5. Check whether 6 ^{n} can end with the digit 0 for any natural number n.**

**Solution:**

No, 6n cannot end with the digit 0 for any natural number n.

To understand why, let's consider the prime factorization of 6:

6 = 2 * 3

In any multiple of 6, there will always be at least one factor of 2 and one factor of 3. Since 2 and 3 are both prime numbers, the product of 2 and 3 will always result in a number ending with the digit 0 only when there are additional factors of 5 present in the multiplication. However, since 6 does not have a factor of 5, no multiple of 6 will end with the digit 0.

Therefore, for any natural number n, 6n cannot end with the digit 0.

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

**Solution:**

To explain why the expressions 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 result in composite numbers, we need to evaluate each expression and analyze the properties of composite numbers.

Expression 1: 7 × 11 × 13 + 13

Calculating the expression:

7 × 11 × 13 + 13 = 1,001 + 13 = 1,014

The resulting number, 1,014, is composite. A composite number is a positive integer greater than 1 that has factors other than 1 and itself. In this case, we can see that 1,014 can be evenly divided by numbers other than 1 and 1,014, such as 2, 3, 6, 9, 18, 53, 106, 159, and 318.

Expression 2: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Calculating the expression:

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5,040 + 5 = 5,045

The resulting number, 5,045, is also composite. It can be divided evenly by numbers other than 1 and 5,045, such as 5, 7, 29, 35, 145, 203, 725, and 1,009.

In both cases, the expressions result in composite numbers because they can be divided evenly by factors other than 1 and the numbers themselves. This distinguishes them from prime numbers, which can only be divided evenly by 1 and the numbers themselves.

**7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?**

**Solution:**

To determine when Sonia and Ravi will meet again at the starting point, we need to find the least common multiple (LCM) of their individual times to complete one round.

Sonia takes 18 minutes to complete one round.

Ravi takes 12 minutes to complete one round.

The LCM of 18 and 12 will give us the time at which they will meet again at the starting point.

Prime factorization of 18: 2 * 3 * 3

Prime factorization of 12: 2 * 2 * 3

To find the LCM, we take the highest power of each prime factor that appears in either number: 2 * 2 * 3 * 3 = 36.

Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.

**Exercise 1.3**

**1. Prove that √**5 **is irrational.**

**Solution:**

To prove that √5 is irrational, we will use a proof by contradiction.

Assume, for the sake of contradiction, that √5 is rational. This means that it can be expressed as a fraction in the form of √5 = a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.

Squaring both sides of the equation, we get 5 = (a^2)/(b^2).

Rearranging the equation, we have a^2 = 5b^2.

From this equation, we can see that a^2 is divisible by 5, which implies that a must also be divisible by 5. Let a = 5k, where k is an integer.

Substituting this back into the equation, we get (5k)^2 = 5b^2.

Simplifying further, we have 25k^2 = 5b^2, which leads to 5k^2 = b^2.

Using the same logic as before, we can conclude that b must also be divisible by 5.

However, this contradicts our initial assumption that a and b have no common factors other than 1. If both a and b are divisible by 5, then they have a common factor of 5, which contradicts our assumption.

Hence, our initial assumption that √5 is rational must be false. Therefore, we can conclude that √5 is irrational.

**2. Prove that 3 + 2√5 + is irrational.**

**Solution:**

To prove that 3 + 2√5 is irrational, we will again use a proof by contradiction.

Assume, for the sake of contradiction, that 3 + 2√5 is rational. This means that it can be expressed as a fraction in the form of 3 + 2√5 = a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.

Rearranging the equation, we have 2√5 = a/b - 3.

Squaring both sides of the equation, we get 4 * 5 = (a^2)/(b^2) - 6a/b + 9.

Simplifying further, we have 20 = (a^2)/(b^2) - 6a/b + 9.

Multiplying both sides by b^2, we get 20b^2 = a^2 - 6ab + 9b^2.

From this equation, we can see that a^2 is divisible by 20, which implies that a must also be divisible by 20. Let a = 20k, where k is an integer.

Substituting this back into the equation, we get 20b^2 = (20k)^2 - 6(20k)b + 9b^2.

Simplifying further, we have 20b^2 = 400k^2 - 120kb + 9b^2.

Rearranging the equation, we have 11b^2 = 120kb - 400k^2.

From this equation, we can see that 11b^2 is divisible by 120, which implies that b must also be divisible by 120. However, this contradicts our initial assumption that a and b have no common factors other than 1.

Therefore, our initial assumption that 3 + 2√5 is rational must be false. Thus, we can conclude that 3 + 2√5 is irrational.

**3. Prove that the following are irrationals:**

**(i) 1/√2**

**(ii) 7√5**

**(iii) 6 + **√**2**

**Solution:**

(i) To prove that 1/√2 is irrational, we will use a proof by contradiction.

Assume, for the sake of contradiction, that 1/√2 is rational. This means that it can be expressed as a fraction in the form of 1/√2 = a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.

Rearranging the equation, we have √2 = b/a.

Squaring both sides of the equation, we get 2 = (b^2)/(a^2).

From this equation, we can see that b^2 is divisible by 2, which implies that b must also be divisible by 2. Let b = 2k, where k is an integer.

Substituting this back into the equation, we get 2 = (4k^2)/(a^2).

Simplifying further, we have 1 = (2k^2)/(a^2).

This implies that a^2 is divisible by 2, which means a must also be divisible by 2.

However, this contradicts our initial assumption that a and b have no common factors other than 1. If both a and b are divisible by 2, then they have a common factor of 2, which contradicts our assumption.

Hence, our initial assumption that 1/√2 is rational must be false. Therefore, we can conclude that 1/√2 is irrational.

(ii) To prove that 7√5 is irrational, we can follow a similar proof by contradiction.

Assume, for the sake of contradiction, that 7√5 is rational. This means that it can be expressed as a fraction in the form of 7√5 = a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.

Squaring both sides of the equation, we get 49 * 5 = (a^2)/(b^2).

This simplifies to 245 = (a^2)/(b^2).

From this equation, we can see that a^2 is divisible by 245, which implies that a must also be divisible by 245. Let a = 245k, where k is an integer.

Substituting this back into the equation, we get 245 = (245k^2)/(b^2).

Simplifying further, we have 1 = (k^2)/(b^2).

This implies that b^2 is equal to k^2, which means b must also be divisible by 245.

However, this contradicts our initial assumption that a and b have no common factors other than 1. If both a and b are divisible by 245, then they have a common factor of 245, which contradicts our assumption.

Hence, our initial assumption that 7√5 is rational must be false. Therefore, we can conclude that 7√5 is irrational.

(iii) To prove that 6 + √2 is irrational, we can again use a proof by contradiction.

Assume, for the sake of contradiction, that 6 + √2 is rational. This means that it can be expressed as a fraction in the form of 6 + √2 = a/b, where a and b are integers with no common factors other than 1, and b is not equal to 0.

Rearranging the equation, we have √2 = (a/b) - 6.

Squaring both sides of the equation, we get 2 = (a^2

)/(b^2) - 12(a/b) + 36.

Multiplying both sides by b^2, we get 2b^2 = a^2 - 12ab + 36b^2.

From this equation, we can see that a^2 is divisible by 2, which implies that a must also be divisible by 2. Let a = 2k, where k is an integer.

Substituting this back into the equation, we get 2b^2 = (2k)^2 - 12(2k)b + 36b^2.

Simplifying further, we have 2b^2 = 4k^2 - 24kb + 36b^2.

Rearranging the equation, we have 34b^2 = 24kb - 4k^2.

From this equation, we can see that 34b^2 is divisible by 24, which implies that b must also be divisible by 24.

However, this contradicts our initial assumption that a and b have no common factors other than 1. If both a and b are divisible by 24, then they have a common factor of 24, which contradicts our assumption.

Hence, our initial assumption that 6 + √2 is rational must be false. Therefore, we can conclude that 6 + √2 is irrational.

#### NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

**Exercise 1.4**

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

**(i)** **13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2 ^{3}5^{2}) (vii) 129/(2^{2}5^{7}7^{5}) (viii) 6/15 (ix) 35/50 (x) 77/210**

**Solution:**

(i) 13/3125: This rational number will have a terminating decimal expansion because the denominator (3125) can be expressed as 5^5, which means it only has factors of 2 and 5.

(ii) 17/8: This rational number will have a non-terminating decimal expansion because the denominator (8) does not have only factors of 2 and 5.

(iii) 64/455: This rational number will have a non-terminating decimal expansion because the denominator (455) does not have only factors of 2 and 5.

(iv) 15/1600: This rational number will have a terminating decimal expansion because the denominator (1600) can be expressed as 2^6 * 5^2, which means it only has factors of 2 and 5.

(v) 29/343: This rational number will have a non-terminating decimal expansion because the denominator (343) does not have only factors of 2 and 5.

(vi) 23/(2352): This rational number will have a non-terminating decimal expansion because the denominator (2352) does not have only factors of 2 and 5.

(vii) 129/(225775): This rational number will have a non-terminating decimal expansion because the denominator (225775) does not have only factors of 2 and 5.

(viii) 6/15: This rational number will have a terminating decimal expansion because the denominator (15) can be expressed as 3 * 5, which means it only has factors of 2 and 5.

(ix) 35/50: This rational number will have a terminating decimal expansion because the denominator (50) can be expressed as 2 * 5^2, which means it only has factors of 2 and 5.

(x) 77/210: This rational number will have a non-terminating decimal expansion because the denominator (210) does not have only factors of 2 and 5.

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

**Solution:**

The rational numbers in Question 1 with terminating decimal expansions are:

(i) 13/3125 = 0.00416

(iv) 15/1600 = 0.009375

(viii) 6/15 = 0.4

(ix) 35/50 = 0.7

**3. 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, p q what can you say about the prime factors of q?**

**(i) 43.123456789**

**(ii) 0.120120012000120000. . .**

**Solution:**

(i) The number 43.123456789 is rational. Since it has a finite decimal expansion, it can be expressed as a fraction in the form p/q. The prime factors of q would depend on the specific calculation of the fraction. To determine the prime factors of q, we would need to find the simplified form of the fraction.

(ii) The number 0.120120012000120000… is irrational. It has a non-repeating, non-terminating decimal expansion. Irrational numbers cannot be expressed as a fraction p/q, where p and q are integers and q is nonzero. Therefore, we cannot say anything about the prime factors of q in this case.