ML Aggarwal Solutions for Class 9 Maths Chapter 1 - Rational and Irrational Numbers

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – Rational and Irrational Numbers are provided here to help students prepare for their exams and score well. This chapter mainly deals with problems based on rational and irrational numbers. To solve tricky problems one should have a thorough acquaintance with the formulas discussed in this chapter. Experts at BYJU’S have formulated the solutions where students can secure good marks in the exams by solving all the questions and cross-checking the answers with the ML Aggarwal Solutions prepared by the experts in Maths. These solutions mainly focus on learning various Mathematical tricks and shortcuts for quick and easy calculations. It also helps boost confidence among students in understanding the concepts covered in this chapter. Students can easily download the pdf consisting of this chapter solutions, which are available in the links provided below.

Chapter 1 – Rational and Irrational Numbers contains five exercises and the

ML Aggarwal Class 9 Solutions present in this page provide solutions to questions related to each exercise present in this chapter.

Download the Pdf of ML Aggarwal Solutions for Class 9 Maths Chapter 1 – Rational and Irrational Numbers

 

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Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 1 – Rational and Irrational Numbers

EXERCISE 1.1

1. Insert a rational number between and 2/9 and 3/8 arrange in descending order.

Solution:

Given:

Rational numbers: 2/9 and 3/8

Let us rationalize the numbers,

By taking LCM for denominators 9 and 8 which is 72.

2/9 = (2×8)/(9×8) = 16/72

3/8 = (3×9)/(8×9) = 27/72

Since,

16/72 < 27/72

So, 2/9 < 3/8

The rational number between 2/9 and 3/8 is

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 1

Hence, 3/8 > 43/144 > 2/9

The descending order of the numbers is 3/8, 43/144, 2/9

2. Insert two rational numbers between 1/3 and 1/4 and arrange in ascending order.

Solution:

Given:

The rational numbers 1/3 and ¼

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 2

= 7/24

Now let us find the rational number between ¼ and 7/24

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 3

= 13/48

So,

The two rational numbers between 1/3 and ¼ are

7/24 and 13/48

Hence, we know that, 1/3 > 7/24 > 13/48 > ¼

The ascending order is as follows: ¼, 13/48, 7/24, 1/3

3. Insert two rational numbers between – 1/3 and – 1/2 and arrange in ascending order.

Solution:

Given:

The rational numbers -1/3 and -1/2

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 4

= -5/12

So, the rational number between -1/3 and -1/2 is -5/12

-1/3 > -5/12 > -1/2

Now, let us find the rational number between -1/3 and -5/12

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 5

= -9/24

= -3/8

So, the rational number between -1/3 and -5/12 is -3/8

-1/3 > -3/8 > -5/12

Hence, the two rational numbers between -1/3 and -1/2 are

-1/3 > -3/8 > -5/12 > -1/2

The ascending is as follows: -1/2, -5/12, -3/8, -1/3

4. Insert three rational numbers between 1/3 and 4/5, and arrange in descending order.

Solution:

Given:

The rational numbers 1/3 and 4/5

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 6

= 17/30

So, the rational number between 1/3 and 4/5 is 17/30

1/3 < 17/30 < 4/5

Now, let us find the rational numbers between 1/3 and 17/30

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 7

= 27/60

So, the rational number between 1/3 and 17/30 is 27/60

1/3 < 27/60 < 17/30

Now, let us find the rational numbers between 17/30 and 4/5

By taking LCM and rationalizing, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 8

= 41/60

So, the rational number between 17/30 and 4/5 is 41/60

17/30 < 41/60 < 4/5

Hence, the three rational numbers between 1/3 and 4/5 are

1/3 < 27/60 < 17/30 < 41/60 < 4/5

The descending order is as follows: 4/5, 41/60, 17/30, 27/60, 1/3

5. Insert three rational numbers between 4 and 4.5.

Solution:

Given:

The rational numbers 4 and 4.5

By rationalizing, we get

= (4 + 4.5)/2

= 8.5 / 2

= 4.25

So, the rational number between 4 and 4.5 is 4.25

4 < 4.25 < 4.5

Now, let us find the rational number between 4 and 4.25

By rationalizing, we get

= (4 + 4.25)/2

= 8.25 / 2

= 4.125

So, the rational number between 4 and 4.25 is 4.125

4 < 4.125 < 4.25

Now, let us find the rational number between 4 and 4.125

By rationalizing, we get

= (4 + 4.125)/2

= 8.125 / 2

= 4.0625

So, the rational number between 4 and 4.125 is 4.0625

4 < 4.0625 < 4.125

Hence, the rational numbers between 4 and 4.5 are

4 < 4.0625 < 4.125 < 4.25 < 4.5

The three rational numbers between 4 and 4.5

4.0625, 4.125, 4.25

6. Find six rational numbers between 3 and 4.

Solution:

Given:

The rational number 3 and 4

So let us find the six rational numbers between 3 and 4,

First rational number between 3 and 4 is

= (3 + 4) / 2

= 7/2

Second rational number between 3 and 7/2 is

= (3 + 7/2) / 2

= (6+7) / (2 × 2) [By taking 2 as LCM]

= 13/4

Third rational number between 7/2 and 4 is

= (7/2 + 4) / 2

= (7+8) / (2 × 2) [By taking 2 as LCM]

= 15/4

Fourth rational number between 3 and 13/4 is

= (3 + 13/4) / 2

= (12+13) / (4 × 2) [By taking 4 as LCM]

= 25/8

Fifth rational number between 13/4 and 7/2 is

= [(13/4) + (7/2)] / 2

= [(13+14)/4] / 2 [By taking 4 as LCM]

= (13 + 14) / (4 × 2)

= 27/8

Sixth rational number between 7/2 and 15/4 is

= [(7/2) + (15/4)] / 2

= [(14 + 15)/4] / 2 [By taking 4 as LCM]

= (14 + 15) / (4 × 2)

= 29/8

Hence, the six rational numbers between 3 and 4 are

25/8, 13/4, 27/8, 7/2, 29/8, 15/4

7. Find five rational numbers between 3/5 and 4/5.

Solution:

Given:

The rational numbers 3/5 and 4/5

Now, let us find the five rational numbers between 3/5 and 4/5

So we need to multiply both numerator and denominator with 5 + 1 = 6

We get,

3/5 = (3 × 6) / (5 × 6) = 18/30

4/5 = (4 × 6) / (5 × 6) = 24/30

Now, we have 18/30 < 19/30 < 20/30 < 21/30 < 22/30 < 23/30 < 24/30

Hence, the five rational numbers between 3/5 and 4/5 are

19/30, 20/30, 21/30, 22/30, 23/30

8. Find ten rational numbers between -2/5 and 1/7.

Solution:

Given:

The rational numbers -2/5 and 1/7

By taking LCM for 5 and 7 which is 35

So, -2/5 = (-2 × 7) / (5 × 7) = -14/35

1/7 = (1 × 5) / (7 × 5) = 5/35

Now, we can insert any10 numbers between -14/35 and 5/35

i.e., -13/35, -12/35, -11/35, -10/35, -9/35, -8/35, -7/35, -6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35

Hence, the ten rational numbers between -2/5 and 1/7 are

-6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35

9. Find six rational numbers between 1/2 and 2/3.

Solution:

Given:

The rational number ½ and 2/3

To make the denominators similar let us take LCM for 2 and 3 which is 6

½ = (1 × 3) / (2 × 3) = 3/6

2/3 = (2 × 2) / (3 × 2) = 4/6

Now, we need to insert six rational numbers, so multiply both numerator and denominator by 6 + 1 = 7

3/6 = (3 × 7) / (6 × 7) = 21/42

4/6 = (4 × 7) / (6 × 7) = 28/42

We know that, 21/42 < 22/42 < 23/42 < 24/42 < 25/42 < 26/42 < 27/42 < 28/42

Hence, the six rational numbers between ½ and 2/3 are

22/42, 23/42, 24/42, 25/42, 26/42, 27/42

EXERCISE 1.2

1. Prove that, √5 is an irrational number.

Solution:

Let us consider √5 be a rational number, then

√5 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

5 = p2 / q2

p2 = 5q2 …. (1)

As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.

So 5 divides p

Now, let p = 5k, where ‘k’ is an integer

Square on both sides, we get

p2 = 25k2

5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]

q2 = 5k2

As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.

So 5 divides q

Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √5 is not a rational number.

√5 is an irrational number.

Hence proved.

2. Prove that, √7 is an irrational number.

Solution:

Let us consider √7 be a rational number, then

√7 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

7 = p2 / q2

p2 = 7q2 …. (1)

As we know, ‘7’ divides 7q2, so ‘7’ divides p2 as well. Hence, ‘7’ is prime.

So 7 divides p

Now, let p = 7k, where ‘k’ is an integer

Square on both sides, we get

p2 = 49k2

7q2 = 49k2 [Since, p2 = 7q2, from equation (1)]

q2 = 7k2

As we know, ‘7’ divides 7k2, so ‘7’ divides q2 as well. But ‘7’ is prime.

So 7 divides q

Thus, p and q have a common factor 7. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √7 is not a rational number.

√7 is an irrational number.

Hence proved.

3. Prove that √6 is an irrational number.

Solution:

Let us consider √6 be a rational number, then

√6 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

6 = p2 / q2

p2 = 6q2 …. (1)

As we know, ‘2’ divides 6q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.

So 2 divides p

Now, let p = 2k, where ‘k’ is an integer

Square on both sides, we get

p2 = 4k2

6q2 = 4k2 [Since, p2 = 6q2, from equation (1)]

3q2 = 2k2

As we know, ‘2’ divides 2k2, so ‘2’ divides 3q2 as well.

‘2’ should either divide 3 or divide q2.

But ‘2’ does not divide 3. ‘2’ divides q2 so ‘2’ is prime.

So 2 divides q

Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √6 is not a rational number.

√6 is an irrational number.

Hence proved.

4. Prove that 1/√11 is an irrational number.

Solution:

Let us consider 1/√11 be a rational number, then

1/√11 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

1/11 = p2 / q2

q2 = 11p2 …. (1)

As we know, ‘11’ divides 11p2, so ‘11’ divides q2 as well. Hence, ‘11’ is prime.

So 11 divides q

Now, let q = 11k, where ‘k’ is an integer

Square on both sides, we get

q2 = 121k2

11p2 = 121k2 [Since, q2 = 11p2, from equation (1)]

p2 = 11k2

As we know, ‘11’ divides 11k2, so ‘11’ divides p2 as well. But ‘11’ is prime.

So 11 divides p

Thus, p and q have a common factor 11. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, 1/√11 is not a rational number.

1/√11 is an irrational number.

Hence proved.

5. Prove that √2 is an irrational number. Hence show that 3 — √2 is an irrational.

Solution:

Let us consider √2 be a rational number, then

√2 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

2 = p2 / q2

p2 = 2q2 …. (1)

As we know, ‘2’ divides 2q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.

So 2 divides p

Now, let p = 2k, where ‘k’ is an integer

Square on both sides, we get

p2 = 4k2

2q2 = 4k2 [Since, p2 = 2q2, from equation (1)]

q2 = 2k2

As we know, ‘2’ divides 2k2, so ‘2’ divides q2 as well. But ‘2’ is prime.

So 2 divides q

Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √2 is not a rational number.

√2 is an irrational number.

Now, let us assume 3 – √2 be a rational number, ‘r’

So, 3 – √2 = r

3 – r = √2

We know that, ‘r’ is rational, ‘3- r’ is rational, so ‘√2’ is also rational.

This contradicts the statement that √2 is irrational.

So, 3- √2 is irrational number.

Hence proved.

6. Prove that, √3 is an irrational number. Hence, show that 2/5×√3 is an irrational number.

Solution:

Let us consider √3 be a rational number, then

√3 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

3 = p2 / q2

p2 = 3q2 …. (1)

As we know, ‘3’ divides 3q2, so ‘3’ divides p2 as well. Hence, ‘3’ is prime.

So 3 divides p

Now, let p = 3k, where ‘k’ is an integer

Square on both sides, we get

p2 = 9k2

3q2 = 9k2 [Since, p2 = 3q2, from equation (1)]

q2 = 3k2

As we know, ‘3’ divides 3k2, so ‘3’ divides q2 as well. But ‘3’ is prime.

So 3 divides q

Thus, p and q have a common factor 3. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √3 is not a rational number.

√3 is an irrational number.

Now, let us assume (2/5)√3 be a rational number, ‘r’

So, (2/5)√3 = r

5r/2 = √3

We know that, ‘r’ is rational, ‘5r/2’ is rational, so ‘√3’ is also rational.

This contradicts the statement that √3 is irrational.

So, (2/5)√3 is irrational number.

Hence proved.

7. Prove that √5 is an irrational number. Hence, show that -3 + 2√5 is an irrational number.

Solution:

Let us consider √5 be a rational number, then

√5 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).

So,

5 = p2 / q2

p2 = 5q2 …. (1)

As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.

So 5 divides p

Now, let p = 5k, where ‘k’ is an integer

Square on both sides, we get

p2 = 25k2

5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]

q2 = 5k2

As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.

So 5 divides q

Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that, √5 is not a rational number.

√5 is an irrational number.

Now, let us assume -3 + 2√5 be a rational number, ‘r’

So, -3 + 2√5 = r

-3 – r = 2√5

(-3 – r)/2 = √5

We know that, ‘r’ is rational, ‘(-3 – r)/2’ is rational, so ‘√5’ is also rational.

This contradicts the statement that √5 is irrational.

So, -3 + 2√5 is irrational number.

Hence proved.

8. Prove that the following numbers are irrational:

(i) 5 +√2

(ii) 3 – 5√3

(iii) 2√3 – 7

(iv) √2 +√5

Solution:

(i) 5 +√2

Now, let us assume 5 + √2 be a rational number, ‘r’

So, 5 + √2 = r

r – 5 = √2

We know that, ‘r’ is rational, ‘r – 5’ is rational, so ‘√2’ is also rational.

This contradicts the statement that √2 is irrational.

So, 5 + √2 is irrational number.

(ii) 3 – 5√3

Now, let us assume 3 – 5√3 be a rational number, ‘r’

So, 3 – 5√3 = r

3 – r = 5√3

(3 – r)/5 = √3

We know that, ‘r’ is rational, ‘(3 – r)/5’ is rational, so ‘√3’ is also rational.

This contradicts the statement that √3 is irrational.

So, 3 – 5√3 is irrational number.

(iii) 2√3 – 7

Now, let us assume 2√3 – 7 be a rational number, ‘r’

So, 2√3 – 7 = r

2√3 = r + 7

√3 = (r + 7)/2

We know that, ‘r’ is rational, ‘(r + 7)/2’ is rational, so ‘√3’ is also rational.

This contradicts the statement that √3 is irrational.

So, 2√3 – 7 is irrational number.

(iv) √2 +√5

Now, let us assume √2 +√5 be a rational number, ‘r’

So, √2 +√5 = r

√5 = r – √2

Square on both sides,

(√5)2 = (r – √2)2

5 = r2 + (√2)2 – 2r√2

5 = r2 + 2 – 2√2r

5 – 2 = r2 – 2√2r

r2 – 3 = 2√2r

(r2 – 3)/2r = √2

We know that, ‘r’ is rational, ‘(r2 – 3)/2r’ is rational, so ‘√2’ is also rational.

This contradicts the statement that √2 is irrational.

So, √2 +√5 is irrational number.

EXERCISE 1.3

1. Locate √10 and √17 on the amber line.

Solution:

√10

√10 = √(9 + 1) = √((3)2 + 12)

Now let us construct:

  • Draw a line segment AB = 3cm.
  • At point A, draw a perpendicular AX and cut off AC = 1cm.
  • Join BC.

BC = √10cm

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 9

√17

√17 = √(16 + 1) = √((4)2 + 12)

Now let us construct:

  • Draw a line segment AB = 4cm.
  • At point A, draw a perpendicular AX and cut off AC = 1cm.
  • Join BC.

BC = √17cm

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 10

2. Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:

(i) 36/100

(ii) 4 1/8

(iii) 2/9

(iv) 2/11

(v) 3/13

(vi) 329/400

Solution:

(i) 36/100

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 11

36/100 = 0.36

It is a terminating decimal.

(ii) 4 1/8

4 1/8 = (4×8 + 1)/8 = 33/8

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 12

33/8 = 4.125

It is a terminating decimal.

(iii) 2/9

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 13

2/9 = 0.222

It is a non-terminating recurring decimal.

(iv) 2/11

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 14

2/11 = 0.181

It is a non-terminating recurring decimal.

(v) 3/13

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 15

3/13 = 0.2317692307

It is a non-terminating recurring decimal.

(vi) 329/400

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 16

329/400 = 0.8225

It is a terminating decimal.

3. Without actually performing the king 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) 23/75

(iv) 6/15

(v) 1258/625

(vi) 77/210

Solution:

We know that, if the denominator of a fraction has only 2 or 5 or both factors, it is a terminating decimal otherwise it is non-terminating repeating decimals.

(i) 13/3125

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 17

3125 = 5 × 5 × 5 × 5 × 5

Prime factor of 3125 = 5, 5, 5, 5, 5 [i.e., in the form of 2n, 5n]

It is a terminating decimal.

(ii) 17/8

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 18

8 = 2 × 2 × 2

Prime factor of 8 = 2, 2, 2 [i.e., in the form of 2n, 5n]

It is a terminating decimal.

(iii) 23/75

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 19

75 = 3 × 5 × 5

Prime factor of 75 = 3, 5, 5

It is a non-terminating repeating decimal.

(iv) 6/15

Let us divide both numerator and denominator by 3

6/15 = (6 ÷ 3) / (15 ÷ 3)

= 2/5

Since the denominator is 5.

It is a terminating decimal.

(v) 1258/625

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 20

625 = 5 × 5 × 5 × 5

Prime factor of 625 = 5, 5, 5, 5 [i.e., in the form of 2n, 5n]

It is a terminating decimal.

(vi) 77/210

Let us divide both numerator and denominator by 7

77/210 = (77 ÷ 7) / (210 ÷ 7)

= 11/30

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 21

30 = 2 × 3 × 5

Prime factor of 30 = 2, 3, 5

It is a non-terminating repeating decimal.

4. Without actually performing the long division, find if 987/10500 will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.

Solution:

Given:

The fraction 987/10500

Let us divide numerator and denominator by 21, we get

987/10500 = (987 ÷ 21) / (10500 ÷ 21)

= 47/500

So,

The prime factors for denominator 500 = 2 × 2 × 5 × 5 × 5

Since it is of the form: 2n, 5n

Hence it is a terminating decimal.

5. Write the decimal expansions of the following numbers which have terminating decimal expansions:

(i) 17/8

(ii) 13/3125

(iii) 7/80

(iv) 6/15

(v) 2²×7/54

(vi) 237/1500

Solution:

(i) 17/8

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 22

Denominator, 8 = 2 × 2 × 2

= 23

It is a terminating decimal.

When we divide 17/8, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 23

17/8 = 2.125

(ii) 13/3125

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 24

3125 = 5 × 5 × 5 × 5 × 5

Prime factor of 3125 = 5, 5, 5, 5, 5 [i.e., in the form of 2n, 5n]

It is a terminating decimal.

When we divide 13/3125, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 25

13/3125 = 0.00416

(iii) 7/80

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 26

80 = 2 × 2 × 2 × 2 × 5

Prime factor of 80 = 24, 51 [i.e., in the form of 2n, 5n]

It is a terminating decimal.

When we divide 7/80, we get

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 27

7/80 = 0.0875

(iv) 6/15

Let us divide both numerator and denominator by 3, we get

6/15 = (6 ÷ 3) / (15 ÷ 3)

= 2/5

Since the denominator is 5,

It is terminating decimal.

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 28

6/15 = 0.4

(v) (2²×7)/54

We know that the denominator is 54

It is a terminating decimal.

(2²×7)/54 = (2 × 2 × 7) / (5 × 5 × 5 × 5)

= 28/625

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 29

28/625 = 0.0448

It is a terminating decimal.

(vi) 237/1500

Let us divide both numerator and denominator by 3, we get

237/1500 = (237 ÷ 3) / (1500 ÷ 3)

= 79/500

Since the denominator is 500,

Its factors are, 500 = 2 × 2 × 5 × 5 × 5

= 22 × 53

It is terminating decimal.

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 30

237/1500= 79/500 = 0.1518

6. Write the denominator of the rational number 257/5000 in the form 2m × 5where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.

Solution:

Given:

The fraction 257/5000

Since the denominator is 5000,

The factors for 5000 are:

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 31

5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5

= 23 × 54

257/5000 = 257/(23 × 54)

It is a terminating decimal.

So,

Let us multiply both numerator and denominator by 2, we get

257/5000 = (257 × 2) / (5000 × 2)

= 514/10000

= 0.0514

7. Write the decimal expansion of 1/7. Hence, write the decimal expression of? 2/7, 3/7, 4/7, 5/7 and 6/7.

Solution:

Given:

The fraction: 1/7

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 32

1/7 = 0.142857142857

Since it is recurring,

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 33

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 34

8. Express the following numbers in the form p/q’. Where p and q are both integers and q≠0;

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 35

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 36

Let x =
ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 37= 0.3333…

Since there is one repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 3.3333…

Now, subtract both the values,

9x = 3

x = 3/9

= 1/3

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 38= 1/3

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 39

Let x =
ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 40= 5.2222…

Since there is one repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 52.2222…

Now, subtract both the values,

9x = 52 – 5

9x = 47

x = 47/9

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 41= 47/9

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 42

Let x = 0.404040

Since there is two repeating digit after the decimal point,

Multiplying by 100 on both sides, we get

100x = 40.404040…

Now, subtract both the values,

99x = 40

x = 40/99

0.404040… = 40/99

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 43

Let x =
ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 44= 0.47777…

Since there is one non-repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 4.7777

Since there is one repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

100x = 47.7777

Now, subtract both the values,

90x = 47 – 4

90x = 43

x = 43/90

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 45= 43/90

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 46

Let x =
ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 47= 0.13434343…

Since there is one non-repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 1.343434

Since there is two repeating digit after the decimal point,

Multiplying by 100 on both sides, we get

1000x = 134.343434

Now, subtract both the values,

990x = 133

x = 133/990

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 48= 133/990

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 49

Let x =
ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 50= 0.001001001…

Since there is three repeating digit after the decimal point,

Multiplying by 1000 on both sides, we get

1000x = 1.001001

Now, subtract both the values,

999x = 1

x = 1/999

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 51= 1/999

9. Classify the following numbers as rational or irrational:

(i) √23

(ii) √225

(iii) 0.3796

(iv) 7.478478

(v) 1.101001000100001…

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 52

Solution:

(i) √23

Since, 23 is not a perfect square,

√23 is an irrational number.

(ii) √225

√225 = √(15)2 = 15

Since, 225 is a perfect square,

√225 is a rational number.

(iii) 0.3796

0.3796 = 3796/1000

Since, the decimal expansion is terminating decimal.

0.3796 is a rational number.

(iv) 7.478478

Let x = 7.478478

Since there is three repeating digit after the decimal point,

Multiplying by 1000 on both sides, we get

1000x = 7478.478478…

Now, subtract both the values,

999x = 7478 – 7

999x = 7471

x = 7471/999

7.478478 = 7471/999

Hence, it is neither terminating nor non-terminating or non-repeating decimal.

7.478478 is an irrational number.

(v) 1.101001000100001…

Since number of zero’s between two consecutive ones are increasing. So it is non-terminating or non-repeating decimal.

1.101001000100001… is an irrational number.

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 53

Let x = 345.0456456

Multiplying by 10 on both sides, we get

10x = 3450.456456

Since there is three repeating digit after the decimal point,

Multiplying by 1000 on both sides, we get

1000x = 3450456.456456…

Now, subtract both the values,

10000x – 10x = 3450456 – 345

9990x = 3450111

x = 3450111/9990

Since, it is non-terminating repeating decimal.

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 54is a rational number.

10. Insert… following.

(i) One irrational number between 1/3 and ½

(ii) One irrational number between -2/5 and ½

(iii) One irrational number between 0 and 0.1

Solution:

(i) One irrational number between 1/3 and ½

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 55

1/3 = 0.333…

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 56

½ = 0.5

So there are infinite irrational numbers between 1/3 and ½.

One irrational number among them can be 0.4040040004…

(ii) One irrational number between -2/5 and ½

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 57

-2/5 = -0.4

ML Aggarwal Solutions for Class 9 Maths Chapter 1 – image 58

½ = 0.5

So there are infinite irrational numbers between -2/5 and ½.

One irrational number among them can be 0.1010010001…

(iii) One irrational number between 0 and 0.1

There are infinite irrational numbers between 0 and 1.

One irrational number among them can be 0.06006000600006…

11. Insert two irrational numbers between 2 and 3.

Solution:

2 is expressed as √4

And 3 is expressed as √9

So, two irrational numbers between 2 and 3 or √4 and √9 are √5, √6

12. Write two irrational numbers between 4/9 and 7/11.

Solution:

4/9 is expressed as 0.4444…

7/11 is expressed as 0.636363…

So, two irrational numbers between 4/9 and 7/11 are 0.4040040004… and 0.6060060006…

13. Find one rational number between √2 and √3.

Solution:

√2 is expressed as 1.4142…

√3 is expressed as 1.7320…

So, one rational number between √2 and √3 is 1.5.

14. Find two rational numbers between √12 and √15.

Solution:

√12 = √(4×3) = 2√3

Since, 12 < 12.25 < 12.96 < 15

So, √12 < √12.25 < √12.96 < √15

Hence, two rational numbers between √12 and √15are [√12.25, √12.96] or [√3.5, √3.6].

15. Insert irrational numbers between √5 and √7.

Solution:

Since, 5 < 6 < 7

So, irrational number between √5 and √7 is √6.

16. Insert two irrational numbers between √3 and √7.

Solution:

Since, 3 < 4 < 5 < 6 < 7

So,

√3 < √4 < √5 < √6 < √7

But √4 = 2, which is a rational number.

So,

Two irrational numbers between √3 and √7 are √5 and √6.

EXERCISE 1.4

1. Simplify the following:

(i) √45 – 3√20 + 4√5

(ii) 3√3 + 2√27 + 7/√3

(iii) 6√5 × 2√5

(iv) 8√15 ÷ 2√3

(v) √24/8 + √54/9

(vi) 3/√8 + 1/√2

Solution:

(i) √45 – 3√20 + 4√5

Let us simplify the expression,

√45 – 3√20 + 4√5

= √(9×5) – 3√(4×5) + 4√5

= 3√5 – 3×2√5 + 4√5

= 3√5 – 6√5 + 4√5

= √5

(ii) 3√3 + 2√27 + 7/√3

Let us simplify the expression,

3√3 + 2√27 + 7/√3

= 3√3 + 2√(9×3) + 7√3/(√3×√3) (by rationalizing)

= 3√3 + (2×3)√3 + 7√3/3

= 3√3 + 6√3 + (7/3) √3

= √3 (3 + 6 + 7/3)

= √3 (9 + 7/3)

= √3 (27+7)/3

= 34/3 √3

(iii) 6√5 × 2√5

Let us simplify the expression,

6√5 × 2√5

= 12 × 5

= 60

(iv) 8√15 ÷ 2√3

Let us simplify the expression,

8√15 ÷ 2√3

= (8 √5 √3) / 2√3

= 4√5

(v) √24/8 + √54/9

Let us simplify the expression,

√24/8 + √54/9

= √(4×6)/8 + √(9×6)/9

= 2√6/8 + 3√6/9

= √6/4 + √6/3

By taking LCM

= (3√6 + 4√6)/12

= 7√6/12

(vi) 3/√8 + 1/√2

Let us simplify the expression,

3/√8 + 1/√2

= 3/2√2 + 1/√2

By taking LCM

= (3 + 2)/(2√2)

= 5/(2√2)

By rationalizing,

= 5√2/(2√2 × 2√2)

= 5√2/(2×2)

= 5√2/4

2. Simplify the following:

(i) (5 + √7) (2 + √5)

(ii) (5 + √5) (5 – √5)

(iii) (√5 + √2)2

(iv) (√3 – √7)2

(v) (√2 + √3) (√5 + √7)

(vi) (4 + √5) (√3 – √7)

Solution:

(i) (5 + √7) (2 + √5)

Let us simplify the expression,

= 5(2 + √5) + √7(2 + √5)

= 10 + 5√5 + 2√7 + √35

(ii) (5 + √5) (5 – √5)

Let us simplify the expression,

By using the formula,

(a)2 – (b)2 = (a + b) (a – b)

So,

= (5)2 – (√5)2

= 25 – 5

= 20

(iii) (√5 + √2)2

Let us simplify the expression,

By using the formula,

(a + b)2 = a2 + b2 + 2ab

(√5 + √2)2 = (√5)2 + (√2)2 + 2√5√2

= 5 + 2 + 2√10

= 7 + 2√10

(iv) (√3 – √7)2

Let us simplify the expression,

By using the formula,

(a – b)2 = a2 + b2 – 2ab

(√3 – √7)2 = (√3)2 + (√7)2 – 2√3√7

= 3 + 7 – 2√21

= 10 – 2√21

(v) (√2 + √3) (√5 + √7)

Let us simplify the expression,

= √2(√5 + √7) + √3(√5 + √7)

= √2×√5 + √2×√7 + √3×√5 + √3×√7

= √10 + √14 + √15 + √21

(vi) (4 + √5) (√3 – √7)

Let us simplify the expression,

= 4(√3 – √7) + √5(√3 – √7)

= 4√3 – 4√7 + √15 – √35

3. If √2 = 1.414, then find the value of

(i) √8 + √50 + √72 + √98

(ii) 3√32 – 2√50 + 4√128 – 20√18

Solution:

(i) √8 + √50 + √72 + √98

Let us simplify the expression,

√8 + √50 + √72 + √98

= √(2×4) + √(2×25) + √(2×36) + √(2×49)

= √2 √4 + √2 √25 + √2 √36 + √2 √49

= 2√2 + 5√2 + 6√2 + 7√2

= 20√2

= 20 × 1.414

= 28.28

(ii) 3√32 – 2√50 + 4√128 – 20√18

Let us simplify the expression,

3√32 – 2√50 + 4√128 – 20√18

= 3√(16×2) – 2√(25×2) + 4√(64×2) – 20√(9×2)

= 3√16 √2 – 2√25 √2 + 4√64 √2 – 20√9 √2

= 3.4√2 – 2.5√2 + 4.8√2 – 20.3√2

= 12√2 – 10√2 + 32√2 – 60√2

= (12 – 10 + 32 – 60) √2

= -26√2

= -26 × 1.414

= -36.764

4. If √3 = 1.732, then find the value of

(i) √27 + √75 + √108 – √243

(ii) 5√12 – 3√48 + 6√75 + 7√108

Solution:

(i) √27 + √75 + √108 – √243

Let us simplify the expression,

√27 + √75 + √108 – √243

= √(9×3) + √(25×3) + √(36×3) – √(81×3)

= √9 √3 + √25 √3 + √36 √3 – √81 √3

= 3√3 + 5√3 + 6√3 – 9√3

= (3 + 5 + 6 – 9) √3

= 5√3

= 5 × 1.732

= 8.660

(ii) 5√12 – 3√48 + 6√75 + 7√108

Let us simplify the expression,

5√12 – 3√48 + 6√75 + 7√108

= 5√(4×3) – 3√(16×3) + 6√(25×3) + 7√(36×3)

= 5√4 √3 – 3√16 √3 + 6√25 √3 + 7√36 √3

= 5.2√3 – 3.4√3 + 6.5√3 + 7.6√3

= 10√3 – 12√3 + 30√3 + 42√3

= (10 – 12 + 30 + 42) √3

= 70√3

= 70 × 1.732

= 121.24

5. State which of the following are rational or irrational decimals.

(i) √(4/9), -3/70, √(7/25), √(16/5)

(ii) -√(2/49), 3/200, √(25/3), -√(49/16)

Solution:

(i) √(4/9), -3/70, √(7/25), √(16/5)

√(4/9) = 2/3

-3/70 = -3/70

√(7/25) = √7/5

√(16/5) = 4/√5

So,

√7/5 and 4/√5 are irrational decimals.

2/3 and -3/70 are rational decimals.

(ii) -√(2/49), 3/200, √(25/3), -√(49/16)

-√(2/49) = -√2/7

3/200 = 3/200

√(25/3) = 5/√3

-√(49/16) = -7/4

So,

-√2/7 and 5/√3 are irrational decimals.

3/200 and -7/4 are rational decimals.

6. State which of the following are rational or irrational decimals.

(i) -3√2

(ii) √(256/81)

(iii) √(27×16)

(iv) √(5/36)

Solution:

(i) -3√2

We know that √2 is an irrational number.

So, -3√2 will also be irrational number.

(ii) √(256/81)

√(256/81) = 16/9 = 4/3

It is a rational number.

(iii) √(27×16)

√(27×16) = √(9×3×16) = 3×4√3 = 12√3

It is an irrational number.

(iv) √(5/36)

√(5/36) = √5/6

It is an irrational number.

7. State which of the following are irrational numbers.

(i) 3 – √(7/25)

(ii) -2/3 + ∛2

(iii) 3/√3

(iv) -2/7 ∛5

(v) (2 – √3) (2 + √3)

(vi) (3 + √5)2

(vii) (2/5 √7)2

(viii) (3 – √6)2

Solution:

(i) 3 – √(7/25)

Let us simplify,

3 – √(7/25) = 3 – √7/√25

= 3 – √7/5

Hence, 3 – √7/5 is an irrational number.

(ii) -2/3 + ∛2

Let us simplify,

-2/3 + ∛2 = -2/3 + 21/3

Since, 2 is not a perfect cube.

Hence it is an irrational number.

(iii) 3/√3

Let us simplify,

By rationalizing, we get

3/√3 = 3√3 /(√3×√3)

= 3√3/3

= √3

Hence, 3/√3 is an irrational number.

(iv) -2/7 ∛5

Let us simplify,

-2/7 ∛5 = -2/7 (5)1/3

Since, 5 is not a perfect cube.

Hence it is an irrational number.

(v) (2 – √3) (2 + √3)

Let us simplify,

By using the formula,

(a + b) (a – b) = (a)2 (b)2

(2 – √3) (2 + √3) = (2)2 – (√3)2

= 4 – 3

= 1

Hence, it is a rational number.

(vi) (3 + √5)2

Let us simplify,

By using (a + b)2 = a2 + b2 + 2ab

(3 + √5)2 = 32 + (√5)2 + 2.3.√5

= 9 + 5 + 6√5

= 14 + 6√5

Hence, it is an irrational number.

(vii) (2/5 √7)2

Let us simplify,

(2/5 √7)2 = (2/5 √7) × (2/5 √7)

= 4/ 25 × 7

= 28/25

Hence it is a rational number.

(viii) (3 – √6)2

Let us simplify,

By using (a – b)2 = a2 + b2 – 2ab

(3 – √6)2 = 32 + (√6)2 – 2.3.√6

= 9 + 6 – 6√6

= 15 – 6√6

Hence it is an irrational number.

8. Prove the following are irrational numbers.

(i) ∛2

(ii) ∛3

(iii) ∜5

Solution:

(i) ∛2

We know that ∛2 = 21/3

Let us consider 21/3 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

21/3 = p/q

2 = p3/q3

p3 = 2q3 ….. (1)

We know that, 2 divides 2q3 then 2 divides p3

So, 2 divides p

Now, let us consider p = 2k, where k is an integer

Substitute the value of p in (1), we get

p3 = 2q3

(2k)3 = 2q3

8k3 = 2q3

4k3 = q3

We know that, 2 divides 4k3 then 2 divides q3

So, 2 divides q

Thus p and q have a common factor ‘2’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∛2 is an irrational number.

(ii) ∛3

We know that ∛3 = 31/3

Let us consider 31/3 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

31/3 = p/q

3 = p3/q3

p3 = 3q3 ….. (1)

We know that, 3 divides 3q3 then 3 divides p3

So, 3 divides p

Now, let us consider p = 3k, where k is an integer

Substitute the value of p in (1), we get

p3 = 3q3

(3k)3 = 3q3

9k3 = 3q3

3k3 = q3

We know that, 3 divides 9k3 then 3 divides q3

So, 3 divides q

Thus p and q have a common factor ‘3’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∛3 is an irrational number.

(iii) ∜5

We know that ∜5 = 51/4

Let us consider 51/4 = p/q, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

51/4 = p/q

5 = p4/q4

P4 = 5q4 ….. (1)

We know that, 5 divides 5q4 then 5 divides p4

So, 5 divides p

Now, let us consider p = 5k, where k is an integer

Substitute the value of p in (1), we get

P4 = 5q4

(5k)4 = 5q4

625k4 = 5q4

125k4 = q4

We know that, 5 divides 125k4 then 5 divides q4

So, 5 divides q

Thus p and q have a common factor ‘5’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, ∜5 is an irrational number.

9. Find the greatest and the smallest real numbers.

(i) 2√3, 3/√2, -√7, √15

(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3

Solution:

(i) 2√3, 3/√2, -√7, √15

Let us simplify each fraction

2√3 = √(4×3) = √12

3/√2 = (3×√2)/(√2×√2) = 3√2/2 = √((9/4)×2) = √(9/2) = √4.5

-√7 = -√7

√15 = √15

So,

The greatest real number = √15

Smallest real number = -√7

(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3

Let us simplify each fraction

-3√2 = -√(9×2) = -√18

9/√5 = (9×√5)/(√5×√5) = 9√5/5 = √((81/25)×5) = √(81/5) = √16.2

-4 = -√16

4/3 √5 = √((16/9)×5) = √(80/9) = √8.88 = √8.8

3/2√3 = √((9/4)×3) = √(27/4) = √6.25

So,

The greatest real number = 9√5

Smallest real number = -3√2

10. Write in ascending order.

(i) 3√2, 2√3, √15, 4

(ii) 3√2, 2√8, 4, √50, 4√3

Solution:

(i) 3√2, 2√3, √15, 4

3√2 = √(9×2) =√18

2√3 = √(4×3) =√12

√15 = √15

4 = √16

Now, let us arrange in ascending order

√12, √15, √16, √18

So,

2√3, √15, 4, 3√2

(ii) 3√2, 2√8, 4, √50, 4√3

3√2 = √(9×2) =√18

2√8 = √(4×8) =√32

4 = √16

√50 = √50

4√3 =√(16×3) = √48

Now, let us arrange in ascending order

√16, √18, √32, √48, √50

So,

4, 3√2, 2√8, 4√3, √50

11. Write in descending order.

(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)

(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7

Solution:

(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)

9/√2 = (9×√2)/(√2×√2) = 9√2/2 = √((81/4)×2) = √(81/2) = √40.5

3/2 √5 = √((9/4)×5) = √(45/4) = √11.25

4√3 = √(16×3) = √48

3√(6/5) = √((9×6)/5) = √(54/5) = √10.8

Now, let us arrange in descending order

√48, √40.5, √11.25, √10.8

So,

4√3, 9/√2, 3/2 √5, 3√(6/5)

(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7

5/√3 = √(25/3) = √8.33

7/3 √2 = √((49/9) ×2) = √98/9 = √10.88

-√3 = -√3

3√5 = √(9×5) =√45

2√7 = √(4×7) = √28

Now, let us arrange in descending order

√45, √28, √10.88.., √8.33.., -√3

So,

3√5, 2√7, 7/3√2, 5/√3, -√3

12. Arrange in ascending order.

∛2, √3, 6√5

Solution:

Here we can express the given expressions as:

∛2 = 21/3

√3 = 31/2

6√5 = 51/6

Let us make the roots common so,

21/3= 2(2× 1/2 × 1/3) = 41/6

31/2 = 3(3× 1/3 × 1/2) = 271/6

51/6 = 51/6

Now, let us arrange in ascending order,

41/6, 51/6, 271/6

So,

21/3, 51/6, 31/2

So,

∛2, 6√5, √3

EXERCISE 1.5

1. Rationalize the following:

(i) 3/4√5

(ii) 5√7 / √3

(iii) 3/(4 – √7)

(iv) 17/(3√2 + 1)

(v) 16/ (√41 – 5)

(vi) 1/ (√7 – √6)

(vii) 1/ (√5 + √2)

(viii) (√2 + √3) / (√2 – √3)

Solution:

(i) 3/4√5

Let us rationalize,

3/4√5 = (3×√5) /(4√5×√5)

= (3√5) / (4×5)

= (3√5) / 20

(ii) 5√7 / √3

Let us rationalize,

5√7 / √3 = (5√7×√3) / (√3×√3)

= 5√21/3

(iii) 3/(4 – √7)

Let us rationalize,

3/(4 – √7) = [3×(4 + √7)] / [(4 – √7) × (4 + √7)]

= 3(4 + √7) / [42 – (√7)2]

= 3(4 + √7) / [16 – 7]

= 3(4 + √7) / 9

= (4 + √7) / 3

(iv) 17/(3√2 + 1)

Let us rationalize,

17/(3√2 + 1) = 17(3√2 – 1) / [(3√2 + 1) (3√2 – 1)]

= 17(3√2 – 1) / [(3√2)2 – 12]

= 17(3√2 – 1) / [9.2 – 1]

= 17(3√2 – 1) / [18 – 1]

= 17(3√2 – 1) / 17

= (3√2 – 1)

(v) 16/ (√41 – 5)

Let us rationalize,

16/ (√41 – 5) = 16(√41 + 5) / [(√41 – 5) (√41 + 5)]

= 16(√41 + 5) / [(√41)2 – 52]

= 16(√41 + 5) / [41 – 25]

= 16(√41 + 5) / [16]

= (√41 + 5)

(vi) 1/ (√7 – √6)

Let us rationalize,

1/ (√7 – √6) = 1(√7 + √6) / [(√7 – √6) (√7 + √6)]

= (√7 + √6) / [(√7)2 – (√6)2]

= (√7 + √6) / [7 – 6]

= (√7 + √6) / 1

= (√7 + √6)

(vii) 1/ (√5 + √2)

Let us rationalize,

1/ (√5 + √2) = 1(√5 – √2) / [(√5 + √2) (√5 – √2)]

= (√5 – √2) / [(√5)2 – (√2)2]

= (√5 – √2) / [5 – 2]

= (√5 – √2) / [3]

= (√5 – √2) /3

(viii) (√2 + √3) / (√2 – √3)

Let us rationalize,

(√2 + √3) / (√2 – √3) = [(√2 + √3) (√2 + √3)] / [(√2 – √3) (√2 + √3)]

= [(√2 + √3)2] / [(√2)2 – (√3)2]

= [2 + 3 + 2√2√3] / [2 – 3]

= [5 + 2√6] / -1

= – (5 + 2√6)

2. Simplify:

(i) (7 + 3√5) / (7 – 3√5)

(ii) (3 – 2√2) / (3 + 2√2)

(iii) (5 – 3√14) / (7 + 2√14)

Solution:

(i) (7 + 3√5) / (7 – 3√5)

Let us rationalize the denominator, we get

(7 + 3√5) / (7 – 3√5) = [(7 + 3√5) (7 + 3√5)] / [(7 – 3√5) (7 + 3√5)]

= [(7 + 3√5)2] / [72 – (3√5)2]

= [72 + (3√5)2 + 2.7. 3√5] / [49 – 9.5]

= [49 + 9.5 + 42√5] / [49 – 45]

= [49 + 45 + 42√5] / [4]

= [94 + 42√5] / 4

= 2[47 + 21√5]/4

= [47 + 21√5]/2

(ii) (3 – 2√2) / (3 + 2√2)

Let us rationalize the denominator, we get

(3 – 2√2) / (3 + 2√2) = [(3 – 2√2) (3 – 2√2)] / [(3 + 2√2) (3 – 2√2)]

= [(3 – 2√2)2] / [32 – (2√2)2]

= [32 + (2√2)2 – 2.3.2√2] / [9 – 4.2]

= [9 + 4.2 – 12√2] / [9 – 8]

= [9 + 8 – 12√2] / 1

= 17 – 12√2

(iii) (5 – 3√14) / (7 + 2√14)

Let us rationalize the denominator, we get

(5 – 3√14) / (7 + 2√14) = [(5 – 3√14) (7 – 2√14)] / [(7 + 2√14) (7 – 2√14)]

= [5(7 – 2√14) – 3√14 (7 – 2√14)] / [72 – (2√14)2]

= [35 – 10√14 – 21√14 + 6.14] / [49 – 4.14]

= [35 – 31√14 + 84] / [49 – 56]

= [119 – 31√14] / [-7]

= -[119 – 31√14] / 7

= [31√14 – 119] / 7

3. Simplify:

[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]

Solution:

Let us simplify individually,

[7√3 / (√10 + √3)]

Let us rationalize the denominator,

7√3 / (√10 + √3) = [7√3(√10 – √3)] / [(√10 + √3) (√10 – √3)]

= [7√3.√10 – 7√3.√3] / [(√10)2 – (√3)2]

= [7√30 – 7.3] / [10 – 3]

= 7[√30 – 3] / 7

= √30 – 3

Now,

[2√5 / (√6 + √5)]

Let us rationalize the denominator, we get

2√5 / (√6 + √5) = [2√5 (√6 – √5)] / [(√6 + √5) (√6 – √5)]

= [2√5.√6 – 2√5.√5] / [(√6)2 – (√5)2]

= [2√30 – 2.5] / [6 – 5]

= [2√30 – 10] / 1

= 2√30 – 10

Now,

[3√2 / (√15 + 3√2)]

Let us rationalize the denominator, we get

3√2 / (√15 + 3√2) = [3√2 (√15 – 3√2)] / [(√15 + 3√2) (√15 – 3√2)]

= [3√2.√15 – 3√2.3√2] / [(√15)2 – (3√2)2]

= [3√30 – 9.2] / [15 – 9.2]

= [3√30 – 18] / [15 – 18]

= 3[√30 – 6] / [-3]

= [√30 – 6] / -1

= 6 – √30

So, according to the question let us substitute the obtained values,

[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]

= (√30 – 3) – (2√30 – 10) – (6 – √30)

= √30 – 3 – 2√30 + 10 – 6 + √30

= 2√30 – 2√30 – 3 + 10 – 6

= 1

4. Simplify:

[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]

Solution:

Let us simplify individually,

[1/(√4 + √5)]

Rationalize the denominator, we get

[1/(√4 + √5)] = [1(√4 – √5)] / [(√4 + √5) (√4 – √5)]

= [(√4 – √5)] / [(√4)2 – (√5)2]

= [(√4 – √5)] / [4 – 5]

= [(√4 – √5)] / -1

= -(√4 – √5)

Now,

[1/(√5 + √6)]

Rationalize the denominator, we get

[1/(√5 + √6)] = [1(√5 – √6)] / [(√5 + √6) (√5 – √6)]

= [(√5 – √6)] / [(√5)2 – (√6)2]

= [(√5 – √6)] / [5 – 6]

= [(√5 – √6)] / -1

= -(√5 – √6)

Now,

[1/(√6 + √7)]

Rationalize the denominator, we get

[1/(√6 + √7)] = [1(√6 – √7)] / [(√6 + √7) (√6 – √7)]

= [(√6 – √7)] / [(√6)2 – (√7)2]

= [(√6 – √7)] / [6 – 7]

= [(√6 – √7)] / -1

= -(√6 – √7)

Now,

[1/(√7 + √8)]

Rationalize the denominator, we get

[1/(√7 + √8)] = [1(√7 – √8)] / [(√7 + √8) (√7 – √8)]

= [(√7 – √8)] / [(√7)2 – (√8)2]

= [(√7 – √8)] / [7 – 8]

= [(√7 – √8)] / -1

= -(√7 – √8)

Now,

[1/(√8 + √9)]

Rationalize the denominator, we get

[1/(√8 + √9)] = [1(√8 – √9)] / [(√8 + √9) (√8 – √9)]

= [(√8 – √9)] / [(√8)2 – (√9)2]

= [(√8 – √9)] / [8 – 9]

= [(√8 – √9)] / -1

= -(√8 – √9)

So, according to the question let us substitute the obtained values,

[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]

= -(√4 – √5) + -(√5 – √6) + -(√6 – √7) + -(√7 – √8) + -(√8 – √9)

= -√4 + √5 – √5 + √6 – √6 + √7 – √7 + √8 – √8 + √9

= -√4 + √9

= -2 + 3

= 1

5. Given, find the value of a and b, if

(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5

(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6

(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5

Solution:

(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5

Let us consider LHS

[3 – √5] / [3 + 2√5]

Rationalize the denominator,

[3 – √5] / [3 + 2√5] = [(3 – √5) (3 – 2√5)] / [(3 + 2√5) (3 – 2√5)]

= [3(3 – 2√5) – √5(3 – 2√5)] / [32 – (2√5)2]

= [9 – 6√5 – 3√5 + 2.5] / [9 – 4.5]

= [9 – 6√5 – 3√5 + 10] / [9 – 20]

= [19 – 9√5] / -11

= -19/11 + 9√5/11

So when comparing with RHS

-19/11 + 9√5/11 = -19/11 + a√5

Hence, value of a = 9/11

(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6

Let us consider LHS

[√2 + √3] / [3√2 – 2√3]

Rationalize the denominator,

[√2 + √3] / [3√2 – 2√3] = [(√2 + √3) (3√2 + 2√3)] / [(3√2 – 2√3) (3√2 + 2√3)]

= [√2(3√2 + 2√3) + √3(3√2 + 2√3)] / [(3√2)2 – (2√3)2]

= [3.2 + 2√2√3 + 3√2√3 + 2.3] / [9.2 – 4.3]

= [6 + 2√6 + 3√6 + 6] / [18 – 12]

= [12 + 5√6] / 6

= 12/6 + 5√6/6

= 2 + 5√6/6

= 2 – (-5√6/6)

So when comparing with RHS

2 – (-5√6/6) = a – b√6

Hence, value of a = 2 and b = -5/6

(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + √5]/[7 – √5]}

Rationalize the denominator,

[7 + √5]/[7 – √5] = [(7 + √5) (7 + √5)] / [(7 – √5) (7 + √5)]

= [(7 + √5)2] / [72 – (√5)2]

= [72 + (√5)2 + 2.7.√5] / [49 – 5]

= [49 + 5 + 14√5] / [44]

= [54 + 14√5] / 44

Now,

{[7 – √5]/[7 + √5]}

Rationalize the denominator,

[7 – √5]/[7 + √5] = (7 – √5) (7 – √5)] / [(7 + √5) (7 – √5)]

= [(7 – √5)2] / [72 – (√5)2]

= [72 + (√5)2 – 2.7.√5] / [49 – 5]

= [49 + 5 – 14√5] / [44]

= [54 – 14√5] / 44

So, according to the question

{[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]}

By substituting the obtained values,

= {[54 + 14√5] / 44} – {[54 – 14√5] / 44}

= [54 + 14√5 – 54 + 14√5]/44

= 28√5/44

= 7√5/11

So when comparing with RHS

7√5/11 = a + 7/11 b√5

Hence, value of a = 0 and b = 1

6. Simplify:

{[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]} = p + q√5

Solution:

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + 3√5] / [3 + √5]}

Rationalize the denominator,

[7 + 3√5] / [3 + √5] = [(7 + 3√5) (3 – √5)] / [(3 + √5) (3 – √5)]

= [7(3 – √5) + 3√5(3 – √5)] / [32 – (√5)2]

= [21 – 7√5 + 9√5 – 3.5] / [9 – 5]

= [21 + 2√5 – 15] / [4]

= [6 + 2√5] / 4

= 2[3 + √5]/4

= [3 + √5] /2

Now,

{[7 – 3√5] / [3 – √5]}

Rationalize the denominator,

[7 – 3√5] / [3 – √5] = [(7 – 3√5) (3 + √5)] / [(3 – √5) (3 + √5)]

= [7(3 + √5) – 3√5(3 + √5)] / [32 – (√5)2]

= [21 + 7√5 – 9√5 – 3.5] / [9 – 5]

= [21 – 2√5 – 15] / 4

= [6 – 2√5]/4

= 2[3 – √5]/4

= [3 – √5]/2

So, according to the question

{[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]}

By substituting the obtained values,

= {[3 + √5] /2} – {[3 – √5] /2}

= [3 + √5 – 3 + √5]/2

= [2√5]/2

= √5

So when comparing with RHS

√5 = p + q√5

Hence, value of p = 0 and q = 1

7. If √2 = 1.414, √3 = 1.732, find

(i) √2/(2 + √2)

(ii) 1/(√3 + √2)

Solution:

(i) √2/(2 + √2)

By rationalizing the denominator,

√2/(2 + √2) = [√2(2 – √2)] / [(2 + √2) (2 – √2)]

= [2√2 – 2] / [22 – (√2)2]

= [2√2 – 2] / [4 – 2]

= 2[√2 – 1] / 2

= √2 – 1

= 1.414 – 1

= 0.414

(ii) 1/(√3 + √2)

By rationalizing the denominator,

1/(√3 + √2) = [1(√3 – √2)] / [(√3 + √2) (√3 – √2)]

= [(√3 – √2)] / [(√3)2 – (√2)2]

= [(√3 – √2)] / [3 – 2]

= [(√3 – √2)] / 1

= (√3 – √2)

= 1.732 – 1.414

= 0.318

8. If a = 2 + √3, find 1/a, (a – 1/a)

Solution:

Given:

a = 2 + √3

So,

1/a = 1/ (2 + √3)

By rationalizing the denominator,

1/ (2 + √3) = [1(2 – √3)] / [(2 + √3) (2 – √3)]

= [(2 – √3)] / [22 – (√3)2]

= [(2 – √3)] / [4 – 3]

= (2 – √3)

Then,

a – 1/a = 2 + √3 – (2 – √3)

= 2 + √3 – 2 + √3

= 2√3

9. Solve:

If x = 1 – √2, find 1/x, (x – 1/x)4

Solution:

Given:

x = 1 – √2

so,

1/x = 1/(1 – √2)

By rationalizing the denominator,

1/ (1 – √2) = [1(1 + √2)] / [(1 – √2) (1 + √2)]

= [(1 + √2)] / [12 – (√2)2]

= [(1 + √2)] / [1 – 2]

= (1 + √2) / -1

= -(1 + √2 )

Then,

(x – 1/x)4 = [1 – √2 – (-1 – √2)]4

= [1 – √2 + 1 + √2]4

= 24

= 16

10. Solve:

If x = 5 – 2√6, find 1/x, (x2 – 1/x2)

Solution:

Given:

x = 5 – 2√6

so,

1/x = 1/(5 – 2√6)

By rationalizing the denominator,

1/(5 – 2√6) = [1(5 + 2√6)] / [(5 – 2√6) (5 + 2√6)]

= [(5 + 2√6)] / [52 – (2√6)2]

= [(5 + 2√6)] / [25 – 4.6]

= [(5 + 2√6)] / [25 – 24]

= (5 + 2√6)

Then,

x + 1/x = 5 – 2√6 + (5 + 2√6)

= 10

Square on both sides we get

(x + 1/x)2 = 102

x2 + 1/x2 + 2x.1/x = 100

x2 + 1/x2 + 2 = 100

x2 + 1/x2 = 100 – 2

= 98

11. If p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5), find the values of

(i) p + q

(ii) p – q

(iii) p2 + q2

(iv) p2 – q2

Solution:

Given:

p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5)

(i) p + q

[(2-√5)/(2+√5)] + [(2+√5)/(2-√5)]

So by rationalizing the denominator, we get

= [(2 – √5)2 + (2 + √5)2] / [22 – (√5)2]

= [4 + 5 – 4√5 + 4 + 5 + 4√5] / [4 – 5]

= [18]/-1

= -18

(ii) p – q

[(2-√5)/(2+√5)] – [(2+√5)/(2-√5)]

So by rationalizing the denominator, we get

= [(2 – √5)2 – (2 + √5)2] / [22 – (√5)2]

= [4 + 5 – 4√5 – (4 + 5 + 4√5)] / [4 – 5]

= [9 – 4√5 – 9 – 4√5] / -1

= [-8√5]/-1

= 8√5

(iii) p2 + q2

We know that (p + q)2 = p2 + q2 + 2pq

So,

p2 + q2 = (p + q)2 – 2pq

pq = [(2-√5)/(2+√5)] × [(2+√5)/(2-√5)]

= 1

p + q = -18

so,

p2 + q2 = (p + q)2 – 2pq

= (-18)2 – 2(1)

= 324 – 2

= 322

(iv) p2 – q2

We know that, p2 – q2 = (p + q) (p – q)

So, by substituting the values

p2 – q2 = (p + q) (p – q)

= (-18) (8√5)

= -144√5

12. If x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1), find

(i) x + y

(ii) xy

Solution:

Given:

x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1)

(i) x + y

= [(√2 – 1)/( √2 + 1)] + [(√2 + 1)/( √2 – 1)]

By rationalizing the denominator,

= [(√2 – 1)2 + (√2 + 1)2] / [(√2)2 – 12]

= [2 + 1 – 2√2 + 2 + 1 + 2√2] / [2 – 1]

= [6] / 1

= 6

(ii) xy

[(√2 – 1)/( √2 + 1)] × [(√2 + 1)/( √2 – 1)]

= 1

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