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 BYJUâ€™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.

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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 = (2Ã—8)/(9Ã—8) = 16/72

3/8 = (3Ã—9)/(8Ã—9) = 27/72

Since,

16/72 < 27/72

So, 2/9 < 3/8

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

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 Â¼

By taking LCM and rationalizing, we get

= 7/24

Now let us find the rational number between Â¼ and 7/24

By taking LCM and rationalizing, we get

= 13/48

So,

The two rational numbers between 1/3 and Â¼ are

7/24 and 13/48

Hence, we know that, 1/3 > 7/24 > 13/48 > Â¼

The ascending order is as follows: Â¼, 13/48, 7/24, 1/3

**3. Insert two rational numbers between â€“ 1/3 and â€“ 1/2 and arrange in ascending order.**

**Solution:**

Given:

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

By taking LCM and rationalizing, we get

= -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

= -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

= 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

= 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

= 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 Ã— 2) [By taking 2 as LCM]

= 13/4

Third rational number between 7/2 and 4 is

= (7/2 + 4) / 2

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

= 15/4

Fourth rational number between 3 and 13/4 is

= (3 + 13/4) / 2

= (12+13) / (4 Ã— 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 Ã— 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 Ã— 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 Ã— 6) / (5 Ã— 6) = 18/30

4/5 = (4 Ã— 6) / (5 Ã— 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 Ã— 7) / (5 Ã— 7) = -14/35

1/7 = (1 Ã— 5) / (7 Ã— 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 Â½ and 2/3

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

Â½ = (1 Ã— 3) / (2 Ã— 3) = 3/6

2/3 = (2 Ã— 2) / (3 Ã— 2) = 4/6

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

3/6 = (3 Ã— 7) / (6 Ã— 7) = 21/42

4/6 = (4 Ã— 7) / (6 Ã— 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 Â½ and 2/3 are

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

EXERCISE 1.2

**1. Prove that, âˆš5 is an irrational number.**

**Solution:**

Let us consider âˆš5 be a rational number, then

âˆš5 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

5 = p^{2} / q^{2}

p^{2} = 5q^{2} â€¦. (1)

As we know, â€˜5â€™ divides 5q^{2}, so â€˜5â€™ divides p^{2} as well. Hence, â€˜5â€™ is prime.

So 5 divides p

Now, let p = 5k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 25k^{2}

5q^{2} = 25k^{2} [Since, p^{2} = 5q^{2}, from equation (1)]

q^{2} = 5k^{2}

As we know, â€˜5â€™ divides 5k^{2}, so â€˜5â€™ divides q^{2} as well. But â€˜5â€™ is prime.

So 5 divides q

Thus, p and q have a common factor 5. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš5 is not a rational number.

âˆš5 is an irrational number.

Hence proved.

**2. Prove that, âˆš7 is an irrational number. **

**Solution:**

Let us consider âˆš7 be a rational number, then

âˆš7 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

7 = p^{2} / q^{2}

p^{2} = 7q^{2} â€¦. (1)

As we know, â€˜7â€™ divides 7q^{2}, so â€˜7â€™ divides p^{2} as well. Hence, â€˜7â€™ is prime.

So 7 divides p

Now, let p = 7k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 49k^{2}

7q^{2} = 49k^{2} [Since, p^{2} = 7q^{2}, from equation (1)]

q^{2} = 7k^{2}

As we know, â€˜7â€™ divides 7k^{2}, so â€˜7â€™ divides q^{2} as well. But â€˜7â€™ is prime.

So 7 divides q

Thus, p and q have a common factor 7. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš7 is not a rational number.

âˆš7 is an irrational number.

Hence proved.

**3. Prove that âˆš6 is an irrational number.**

**Solution:**

Let us consider âˆš6 be a rational number, then

âˆš6 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

6 = p^{2} / q^{2}

p^{2} = 6q^{2} â€¦. (1)

As we know, â€˜2â€™ divides 6q^{2}, so â€˜2â€™ divides p^{2} as well. Hence, â€˜2â€™ is prime.

So 2 divides p

Now, let p = 2k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 4k^{2}

6q^{2} = 4k^{2} [Since, p^{2} = 6q^{2}, from equation (1)]

3q^{2} = 2k^{2}

As we know, â€˜2â€™ divides 2k^{2}, so â€˜2â€™ divides 3q^{2} as well.

â€˜2â€™ should either divide 3 or divide q^{2}.

But â€˜2â€™ does not divide 3. â€˜2â€™ divides q^{2} so â€˜2â€™ is prime.

So 2 divides q

Thus, p and q have a common factor 2. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš6 is not a rational number.

âˆš6 is an irrational number.

Hence proved.

**4. Prove that 1/âˆš11 is anÂ irrationalÂ number.**

**Solution:**

Let us consider 1/âˆš11 be a rational number, then

1/âˆš11 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

1/11 = p^{2} / q^{2}

q^{2} = 11p^{2} â€¦. (1)

As we know, â€˜11â€™ divides 11p^{2}, so â€˜11â€™ divides q^{2} as well. Hence, â€˜11â€™ is prime.

So 11 divides q

Now, let q = 11k, where â€˜kâ€™ is an integer

Square on both sides, we get

q^{2} = 121k^{2}

11p^{2} = 121k^{2} [Since, q^{2} = 11p^{2}, from equation (1)]

p^{2} = 11k^{2}

As we know, â€˜11â€™ divides 11k^{2}, so â€˜11â€™ divides p^{2} as well. But â€˜11â€™ is prime.

So 11 divides p

Thus, p and q have a common factor 11. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, 1/âˆš11 is not a rational number.

1/âˆš11 is an irrational number.

Hence proved.

**5. Prove that âˆš2 is an irrational number. Hence show that 3 â€” âˆš2 is an irrational.**

**Solution:**

Let us consider âˆš2 be a rational number, then

âˆš2 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

2 = p^{2} / q^{2}

p^{2} = 2q^{2} â€¦. (1)

As we know, â€˜2â€™ divides 2q^{2}, so â€˜2â€™ divides p^{2} as well. Hence, â€˜2â€™ is prime.

So 2 divides p

Now, let p = 2k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 4k^{2}

2q^{2} = 4k^{2} [Since, p^{2} = 2q^{2}, from equation (1)]

q^{2} = 2k^{2}

As we know, â€˜2â€™ divides 2k^{2}, so â€˜2â€™ divides q^{2} as well. But â€˜2â€™ is prime.

So 2 divides q

Thus, p and q have a common factor 2. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš2 is not a rational number.

âˆš2 is an irrational number.

Now, let us assume 3 – âˆš2 be a rational number, â€˜râ€™

So, 3 – âˆš2 = r

3 â€“ r = âˆš2

We know that, â€˜râ€™ is rational, â€˜3- râ€™ is rational, so â€˜âˆš2â€™ is also rational.

This contradicts the statement that âˆš2 is irrational.

So, 3- âˆš2 is irrational number.

Hence proved.

**6. Prove that, âˆš3 is an irrational number. Hence, show that 2/5Ã—âˆš3 is an irrational number. **

**Solution:**

Let us consider âˆš3 be a rational number, then

âˆš3 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

3 = p^{2} / q^{2}

p^{2} = 3q^{2} â€¦. (1)

As we know, â€˜3â€™ divides 3q^{2}, so â€˜3â€™ divides p^{2} as well. Hence, â€˜3â€™ is prime.

So 3 divides p

Now, let p = 3k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 9k^{2}

3q^{2} = 9k^{2} [Since, p^{2} = 3q^{2}, from equation (1)]

q^{2} = 3k^{2}

As we know, â€˜3â€™ divides 3k^{2}, so â€˜3â€™ divides q^{2} as well. But â€˜3â€™ is prime.

So 3 divides q

Thus, p and q have a common factor 3. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš3 is not a rational number.

âˆš3 is an irrational number.

Now, let us assume (2/5)âˆš3 be a rational number, â€˜râ€™

So, (2/5)âˆš3 = r

5r/2 = âˆš3

We know that, â€˜râ€™ is rational, â€˜5r/2â€™ is rational, so â€˜âˆš3â€™ is also rational.

This contradicts the statement that âˆš3 is irrational.

So, (2/5)âˆš3 is irrational number.

Hence proved.

**7. Prove that âˆš5 is an irrational number. Hence, show that -3 + 2âˆš5 is an irrational number.**

**Solution:**

Let us consider âˆš5 be a rational number, then

âˆš5 = p/q, where â€˜pâ€™ and â€˜qâ€™ are integers, q **â‰ ** 0 and p, q have no common factors (except 1).

So,

5 = p^{2} / q^{2}

p^{2} = 5q^{2} â€¦. (1)

As we know, â€˜5â€™ divides 5q^{2}, so â€˜5â€™ divides p^{2} as well. Hence, â€˜5â€™ is prime.

So 5 divides p

Now, let p = 5k, where â€˜kâ€™ is an integer

Square on both sides, we get

p^{2} = 25k^{2}

5q^{2} = 25k^{2} [Since, p^{2} = 5q^{2}, from equation (1)]

q^{2} = 5k^{2}

As we know, â€˜5â€™ divides 5k^{2}, so â€˜5â€™ divides q^{2} as well. But â€˜5â€™ is prime.

So 5 divides q

Thus, p and q have a common factor 5. This statement contradicts that â€˜pâ€™ and â€˜qâ€™ has no common factors (except 1).

We can say that, âˆš5 is not a rational number.

âˆš5 is an irrational number.

Now, let us assume -3 + 2âˆš5 be a rational number, â€˜râ€™

So, -3 + 2âˆš5 = r

-3 – r = 2âˆš5

(-3 – r)/2 = âˆš5

We know that, â€˜râ€™ is rational, â€˜(-3 – r)/2â€™ is rational, so â€˜âˆš5â€™ is also rational.

This contradicts the statement that âˆš5 is irrational.

So, -3 + 2âˆš5 is irrational number.

Hence proved.

**8. Prove that the following numbers are irrational:**

**(i) 5 +âˆš2**

**(ii) 3 – 5âˆš3**

**(iii) 2âˆš3 – 7**

**(iv) âˆš2 +âˆš5**

**Solution:**

**(i) **5 +âˆš2

Now, let us assume 5 + âˆš2 be a rational number, â€˜râ€™

So, 5 + âˆš2 = r

r – 5 = âˆš2

We know that, â€˜râ€™ is rational, â€˜r – 5â€™ is rational, so â€˜âˆš2â€™ is also rational.

This contradicts the statement that âˆš2 is irrational.

So, 5 + âˆš2 is irrational number.

**(ii) **3 – 5âˆš3

Now, let us assume 3 – 5âˆš3 be a rational number, â€˜râ€™

So, 3 – 5âˆš3 = r

3 – r = 5âˆš3

(3 – r)/5 = âˆš3

We know that, â€˜râ€™ is rational, â€˜(3 – r)/5â€™ is rational, so â€˜âˆš3â€™ is also rational.

This contradicts the statement that âˆš3 is irrational.

So, 3 – 5âˆš3 is irrational number.

**(iii) **2âˆš3 – 7

Now, let us assume 2âˆš3 â€“ 7 be a rational number, â€˜râ€™

So, 2âˆš3 â€“ 7 = r

2âˆš3 = r + 7

âˆš3 = (r + 7)/2

We know that, â€˜râ€™ is rational, â€˜(r + 7)/2â€™ is rational, so â€˜âˆš3â€™ is also rational.

This contradicts the statement that âˆš3 is irrational.

So, 2âˆš3 â€“ 7 is irrational number.

**(iv) **âˆš2 +âˆš5

Now, let us assume âˆš2 +âˆš5 be a rational number, â€˜râ€™

So, âˆš2 +âˆš5 = r

âˆš5 = r – âˆš2

Square on both sides,

(âˆš5)^{2} = (r – âˆš2)^{2}

5 = r^{2} + (âˆš2)^{2} â€“ 2râˆš2

5 = r^{2} + 2 – 2âˆš2r

5 â€“ 2 = r^{2} – 2âˆš2r

r^{2} â€“ 3 = 2âˆš2r

(r^{2} – 3)/2r = âˆš2

We know that, â€˜râ€™ is rational, â€˜(r^{2} – 3)/2râ€™ is rational, so â€˜âˆš2â€™ is also rational.

This contradicts the statement that âˆš2 is irrational.

So, âˆš2 +âˆš5 is irrational number.

EXERCISE 1.3

**1. Locate âˆš10 and âˆš17 on the amber line.**

**Solution:**

âˆš10

âˆš10 = âˆš(9 + 1) = âˆš((3)^{2} + 1^{2})

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 = âˆš10cm

âˆš17

âˆš17 = âˆš(16 + 1) = âˆš((4)^{2} + 1^{2})

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 = âˆš17cm

**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

36/100 = 0.36

It is a terminating decimal.

**(ii) **4 1/8

4 1/8 = (4Ã—8 + 1)/8 = 33/8

33/8 = 4.125

It is a terminating decimal.

**(iii) **2/9

2/9 = 0.222

It is a non-terminating recurring decimal.

**(iv) **2/11

2/11 = 0.181

It is a non-terminating recurring decimal.

**(v) **3/13

3/13 = 0.2317692307

It is a non-terminating recurring decimal.

**(vi) **329/400

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

3125 = 5 Ã— 5 Ã— 5 Ã— 5 Ã— 5

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

It is a terminating decimal.

**(ii) **17/8

8 = 2 Ã— 2 Ã— 2

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

It is a terminating decimal.

**(iii) **23/75

75 = 3 Ã— 5 Ã— 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 **Ã· **3) / (15 **Ã· **3)

= 2/5

Since the denominator is 5.

It is a terminating decimal.

**(v) **1258/625

625 = 5 Ã— 5 Ã— 5 Ã— 5

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

It is a terminating decimal.

**(vi) **77/210

Let us divide both numerator and denominator by 7

77/210 = (77 **Ã· **7) / (210 **Ã· **7)

= 11/30

30 = 2 Ã— 3 Ã— 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 orÂ 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 **Ã· **21) / (10500 **Ã· **21)

= 47/500

So,

The prime factors for denominator 500 = 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5

Since it is of the form: 2^{n}, 5^{n}

Hence it is a terminating decimal.

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

**(i) 17/8**

**(ii) 13/3125**

**(iii) 7/80**

**(iv) 6/15**

**(v) 2Â²Ã—7/5 ^{4}**

**(vi) 237/1500**

**Solution:**

**(i) **17/8

Denominator, 8 = 2 Ã— 2 Ã— 2

= 2^{3}

It is a terminating decimal.

When we divide 17/8, we get

17/8 = 2.125

**(ii) **13/3125

3125 = 5 Ã— 5 Ã— 5 Ã— 5 Ã— 5

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

It is a terminating decimal.

When we divide 13/3125, we get

13/3125 = 0.00416

**(iii) **7/80

80 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5

Prime factor of 80 = 2^{4}, 5^{1} [i.e., in the form of 2^{n}, 5^{n}]

It is a terminating decimal.

When we divide 7/80, we get

7/80 = 0.0875

**(iv) **6/15

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

6/15 = (6 **Ã· **3) / (15 **Ã· **3)

= 2/5

Since the denominator is 5,

It is terminating decimal.

6/15 = 0.4

**(v) (**2Â²Ã—7)/5^{4}

We know that the denominator is 5^{4}

It is a terminating decimal.

**(**2Â²Ã—7)/5^{4} = (2 Ã— 2 Ã— 7) / (5 Ã— 5 Ã— 5 Ã— 5)

= 28/625

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 **Ã· **3) / (1500 **Ã· **3)

= 79/500

Since the denominator is 500,

Its factors are, 500 = 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5

= 2^{2} Ã— 5^{3}

It is terminating decimal.

237/1500= 79/500 = 0.1518

**6. Write the denominator of the rationalÂ numberÂ 257/5000 in the formÂ 2 ^{m }Ã— 5^{nÂ }where m, n isÂ 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:

5000 = 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5 Ã— 5

= 2^{3} Ã— 5^{4}

257/5000 = 257/(2^{3} Ã— 5^{4})

It is a terminating decimal.

So,

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

257/5000 = (257 Ã—** **2) / (5000 Ã—** **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

1/7 = 0.142857142857

Since it is recurring,

**8. Express the following numbers in the form p/qâ€™. Where p and q are both integers andÂ qâ‰ 0;**

**Solution:**

Let x =

= 0.3333â€¦

Since there is one repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 3.3333â€¦

Now, subtract both the values,

9x = 3

x = 3/9

= 1/3

= 1/3

Let x =

= 5.2222â€¦

Since there is one repeating digit after the decimal point,

Multiplying by 10 on both sides, we get

10x = 52.2222â€¦

Now, subtract both the values,

9x = 52 â€“ 5

9x = 47

x = 47/9

= 47/9

Let x = 0.404040

Since there is two repeating digit after the decimal point,

Multiplying by 100 on both sides, we get

100x = 40.404040â€¦

Now, subtract both the values,

99x = 40

x = 40/99

0.404040â€¦ = 40/99

Let x =

= 0.47777â€¦

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 â€“ 4

90x = 43

x = 43/90

= 43/90

Let x =

= 0.13434343â€¦

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

= 133/990

Let x =

= 0.001001001â€¦

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

= 1/999

**9. ClassifyÂ the following numbers as rational or irrational:**

**(i) âˆš23**

**(ii) âˆš225**

**(iii) 0.3796**

**(iv) 7.478478**

**(v) 1.101001000100001â€¦**

**Solution:**

**(i) **âˆš23

Since, 23 is not a perfect square,

âˆš23 is an irrational number.

**(ii) **âˆš225

âˆš225 = âˆš(15)^{2} = 15

Since, 225 is a perfect square,

âˆš225 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 = 7478.478478â€¦

Now, subtract both the values,

999x = 7478 â€“ 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) **1.101001000100001â€¦

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

1.101001000100001â€¦ is an irrational number.

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 = 3450456.456456â€¦

Now, subtract both the values,

10000x â€“ 10x = 3450456 â€“ 345

9990x = 3450111

x = 3450111/9990

Since, it is non-terminating repeating decimal.

is a rational number.

**10. Insertâ€¦ following.**

**(i) One irrational number between 1/3 and Â½**

**(ii) One irrational number between -2/5 and Â½ **

**(iii) One irrational number between 0 and 0.1**

**Solution:**

**(i) **One irrational number between 1/3 and Â½

1/3 = 0.333â€¦

Â½ = 0.5

So there are infinite irrational numbers between 1/3 and Â½.

One irrational number among them can be 0.4040040004â€¦

**(ii) **One irrational number between -2/5 and Â½

-2/5 = -0.4

Â½ = 0.5

So there are infinite irrational numbers between -2/5 and Â½.

One irrational number among them can be 0.1010010001â€¦

**(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 0.06006000600006â€¦

**11. Insert two irrational numbers between 2 and 3.**

**Solution:**

2 is expressed as âˆš4

And 3 is expressed as âˆš9

So, two irrational numbers between 2 and 3 or âˆš4 and âˆš9 are âˆš5, âˆš6

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

**Solution:**

4/9 is expressed as 0.4444â€¦

7/11 is expressed as 0.636363â€¦

So, two irrational numbers between 4/9 and 7/11 are 0.4040040004â€¦ and 0.6060060006â€¦

**13. Find one rational number between âˆš2 and âˆš3.**

**Solution:**

âˆš2 is expressed as 1.4142â€¦

âˆš3 is expressed as 1.7320â€¦

So, one rational number between âˆš2 and âˆš3 is 1.5.

**14. Find two rational numbers between âˆš12 and âˆš15.**

**Solution:**

âˆš12 = âˆš(4Ã—3) = 2âˆš3

Since, 12 < 12.25 < 12.96 < 15

So, âˆš12 < âˆš12.25 < âˆš12.96 < âˆš15

Hence, two rational numbers between âˆš12 and âˆš15are [âˆš12.25, âˆš12.96] or [âˆš3.5, âˆš3.6].

**15. Insert irrational numbers between âˆš5 and âˆš7.**

**Solution:**

Since, 5 < 6 < 7

So, irrational number between âˆš5 and âˆš7 is âˆš6.

**16. Insert two irrational numbers between âˆš3 and âˆš7.**

**Solution:**

Since, 3 < 4 < 5 < 6 < 7

So,

âˆš3 < âˆš4 < âˆš5 < âˆš6 < âˆš7

But âˆš4 = 2, which is a rational number.

So,

Two irrational numbers between âˆš3 and âˆš7 are âˆš5 and âˆš6.

EXERCISE 1.4

**1. Simplify the following:**

**(i) âˆš45 – 3âˆš20 + 4âˆš5**

**(ii) 3âˆš3 + 2âˆš27 + 7/âˆš3**

**(iii) 6âˆš5 Ã— 2âˆš5 **

**(iv) 8âˆš15 Ã· 2âˆš3**

**(v) âˆš24/8 + âˆš54/9**

**(vi) 3/âˆš8 + 1/âˆš2**

**Solution:**

**(i) **âˆš45 – 3âˆš20 + 4âˆš5

Let us simplify the expression,

âˆš45 – 3âˆš20 + 4âˆš5

= âˆš(9Ã—5) – 3âˆš(4Ã—5) + 4âˆš5

= 3âˆš5 – 3Ã—2âˆš5 + 4âˆš5

= 3âˆš5 – 6âˆš5 + 4âˆš5

= âˆš5

**(ii) **3âˆš3 + 2âˆš27 + 7/âˆš3

Let us simplify the expression,

3âˆš3 + 2âˆš27 + 7/âˆš3

= 3âˆš3 + 2âˆš(9Ã—3) + 7âˆš3/(âˆš3Ã—âˆš3) (by rationalizing)

= 3âˆš3 + (2Ã—3)âˆš3 + 7âˆš3/3

= 3âˆš3 + 6âˆš3 + (7/3) âˆš3

= âˆš3 (3 + 6 + 7/3)

= âˆš3 (9 + 7/3)

= âˆš3 (27+7)/3

= 34/3 âˆš3

**(iii) **6âˆš5 Ã— 2âˆš5

Let us simplify the expression,

6âˆš5 Ã— 2âˆš5

= 12 Ã— 5

= 60

**(iv) **8âˆš15 Ã· 2âˆš3

Let us simplify the expression,

8âˆš15 Ã· 2âˆš3

= (8 âˆš5 âˆš3) / 2âˆš3

= 4âˆš5

**(v) **âˆš24/8 + âˆš54/9

Let us simplify the expression,

âˆš24/8 + âˆš54/9

= âˆš(4Ã—6)/8 + âˆš(9Ã—6)/9

= 2âˆš6/8 + 3âˆš6/9

= âˆš6/4 + âˆš6/3

By taking LCM

= (3âˆš6 + 4âˆš6)/12

= 7âˆš6/12

**(vi) **3/âˆš8 + 1/âˆš2

Let us simplify the expression,

3/âˆš8 + 1/âˆš2

= 3/2âˆš2 + 1/âˆš2

By taking LCM

= (3 + 2)/(2âˆš2)

= 5/(2âˆš2)

By rationalizing,

= 5âˆš2/(2âˆš2 Ã— 2âˆš2)

= 5âˆš2/(2Ã—2)

= 5âˆš2/4

**2. Simplify the following:**

**(i) (5 + âˆš7) (2 + âˆš5)**

**(ii) (5 + âˆš5) (5 – âˆš5)**

**(iii) (âˆš5 + âˆš2) ^{2}**

**(iv) (âˆš3 – âˆš7) ^{2}**

**(v) (âˆš2 + âˆš3) (âˆš5 + âˆš7)**

**(vi) (4 + âˆš5) (âˆš3 – âˆš7)**

**Solution:**

**(i) **(5 + âˆš7) (2 + âˆš5)

Let us simplify the expression,

= 5(2 + âˆš5) + âˆš7(2 + âˆš5)

= 10 + 5âˆš5 + 2âˆš7 + âˆš35

**(ii) **(5 + âˆš5) (5 – âˆš5)

Let us simplify the expression,

By using the formula,

(a)^{2} â€“ (b)^{2} = (a + b) (a – b)

So,

= (5)^{2} â€“ (âˆš5)^{2}

= 25 â€“ 5

= 20

**(iii) **(âˆš5 + âˆš2)^{2}

Let us simplify the expression,

By using the formula,

(a + b)^{2} = a^{2} + b^{2} + 2ab

(âˆš5 + âˆš2)^{2}** =** (âˆš5)^{2} + (âˆš2)^{2} + 2âˆš5âˆš2

= 5 + 2 + 2âˆš10

= 7 + 2âˆš10

**(iv) **(âˆš3 – âˆš7)^{2}

Let us simplify the expression,

By using the formula,

(a – b)^{2} = a^{2} + b^{2} – 2ab

(âˆš3 – âˆš7)^{2}** =** (âˆš3)^{2} + (âˆš7)^{2} – 2âˆš3âˆš7

= 3 + 7 – 2âˆš21

= 10 – 2âˆš21

**(v) **(âˆš2 + âˆš3) (âˆš5 + âˆš7)

Let us simplify the expression,

= âˆš2(âˆš5 + âˆš7) + âˆš3(âˆš5 + âˆš7)

= âˆš2Ã—âˆš5 + âˆš2Ã—âˆš7 + âˆš3Ã—âˆš5 + âˆš3Ã—âˆš7

**= **âˆš10 + âˆš14 + âˆš15 + âˆš21

**(vi) **(4 + âˆš5) (âˆš3 – âˆš7)

Let us simplify the expression,

= 4(âˆš3 – âˆš7) + âˆš5(âˆš3 – âˆš7)

= 4âˆš3 – 4âˆš7 + âˆš15 – âˆš35

**3. If âˆš2 = 1.414, then find the value of**

**(i) âˆš8 + âˆš50 + âˆš72 + âˆš98**

**(ii) 3âˆš32 – 2âˆš50 + 4âˆš128 – 20âˆš18**

**Solution:**

**(i) **âˆš8 + âˆš50 + âˆš72 + âˆš98

Let us simplify the expression,

âˆš8 + âˆš50 + âˆš72 + âˆš98

= âˆš(2Ã—4) + âˆš(2Ã—25) + âˆš(2Ã—36) + âˆš(2Ã—49)

= âˆš2 âˆš4 + âˆš2 âˆš25 + âˆš2 âˆš36 + âˆš2 âˆš49

= 2âˆš2 + 5âˆš2 + 6âˆš2 + 7âˆš2

= 20âˆš2

= 20 Ã— 1.414

= 28.28

**(ii) **3âˆš32 – 2âˆš50 + 4âˆš128 – 20âˆš18

Let us simplify the expression,

3âˆš32 – 2âˆš50 + 4âˆš128 – 20âˆš18

= 3âˆš(16Ã—2) – 2âˆš(25Ã—2) + 4âˆš(64Ã—2) – 20âˆš(9Ã—2)

= 3âˆš16 âˆš2 – 2âˆš25 âˆš2 + 4âˆš64 âˆš2 – 20âˆš9 âˆš2

= 3.4âˆš2 â€“ 2.5âˆš2 + 4.8âˆš2 â€“ 20.3âˆš2

= 12âˆš2 – 10âˆš2 + 32âˆš2 – 60âˆš2

= (12 â€“ 10 + 32 – 60) âˆš2

= -26âˆš2

= -26 Ã— 1.414

= -36.764

**4. If âˆš3 = 1.732, then find the value of**

**(i) âˆš27 + âˆš75 + âˆš108 – âˆš243**

**(ii) 5âˆš12 – 3âˆš48 + 6âˆš75 + 7âˆš108**

**Solution:**

**(i) **âˆš27 + âˆš75 + âˆš108 – âˆš243

Let us simplify the expression,

âˆš27 + âˆš75 + âˆš108 – âˆš243

= âˆš(9Ã—3) + âˆš(25Ã—3) + âˆš(36Ã—3) – âˆš(81Ã—3)

= âˆš9 âˆš3 + âˆš25 âˆš3 + âˆš36 âˆš3 – âˆš81 âˆš3

= 3âˆš3 + 5âˆš3 + 6âˆš3 – 9âˆš3

= (3 + 5 + 6 – 9) âˆš3

= 5âˆš3

= 5 Ã— 1.732

= 8.660

**(ii) **5âˆš12 – 3âˆš48 + 6âˆš75 + 7âˆš108

Let us simplify the expression,

5âˆš12 – 3âˆš48 + 6âˆš75 + 7âˆš108

= 5âˆš(4Ã—3) – 3âˆš(16Ã—3) + 6âˆš(25Ã—3) + 7âˆš(36Ã—3)

= 5âˆš4 âˆš3 – 3âˆš16 âˆš3 + 6âˆš25 âˆš3 + 7âˆš36 âˆš3

= 5.2âˆš3 â€“ 3.4âˆš3 + 6.5âˆš3 + 7.6âˆš3

= 10âˆš3 – 12âˆš3 + 30âˆš3 + 42âˆš3

= (10 â€“ 12 + 30 + 42) âˆš3

= 70âˆš3

= 70 Ã— 1.732

= 121.24

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

**(i) âˆš(4/9), -3/70, âˆš(7/25), âˆš(16/5)**

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

**Solution:**

**(i) **âˆš(4/9), -3/70, âˆš(7/25), âˆš(16/5)

âˆš(4/9) = 2/3

-3/70 = -3/70

âˆš(7/25) = âˆš7/5

âˆš(16/5) = 4/âˆš5

So,

âˆš7/5 and 4/âˆš5 are irrational decimals.

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

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

-âˆš(2/49) = -âˆš2/7

3/200 = 3/200

âˆš(25/3) = 5/âˆš3

-âˆš(49/16) = -7/4

So,

-âˆš2/7 and 5/âˆš3 are irrational decimals.

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

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

**(i) -3âˆš2**

**(ii) âˆš(256/81)**

**(iii) âˆš(27Ã—16)**

**(iv) âˆš(5/36)**

**Solution:**

**(i) **-3âˆš2

We know that âˆš2 is an irrational number.

So, -3âˆš2 will also be irrational number.

**(ii) **âˆš(256/81)

âˆš(256/81) = 16/9 = 4/3

It is a rational number.

**(iii) **âˆš(27Ã—16)

âˆš(27Ã—16) = âˆš(9Ã—3Ã—16) = 3Ã—4âˆš3 = 12âˆš3

It is an irrational number.

**(iv) **âˆš(5/36)

âˆš(5/36) = âˆš5/6

It is an irrational number.

**7. State which of the following are irrational numbers.**

**(i) 3 – âˆš(7/25)**

**(ii) -2/3 + âˆ›2**

**(iii) 3/âˆš3 **

**(iv) -2/7 âˆ›5**

**(v) (2 – âˆš3) (2 + âˆš3)**

**(vi) (3 + âˆš5) ^{2} **

**(vii) (2/5 âˆš7) ^{2} **

**(viii) (3 – âˆš6) ^{2}**

**Solution:**

**(i) **3 – âˆš(7/25)

Let us simplify,

3 – âˆš(7/25) = 3 – âˆš7/âˆš25

= 3 – âˆš7/5

Hence, 3 – âˆš7/5 is an irrational number.

**(ii) **-2/3 + âˆ›2

Let us simplify,

-2/3 + âˆ›2 = -2/3 + 2^{1/3}

Since, 2 is not a perfect cube.

Hence it is an irrational number.

**(iii) **3/âˆš3** **

Let us simplify,

By rationalizing, we get

3/âˆš3 = 3âˆš3** **/(âˆš3Ã—âˆš3)

= 3âˆš3/3

= âˆš3

Hence, 3/âˆš3 is an irrational number.

**(iv) **-2/7 âˆ›5

Let us simplify,

-2/7 âˆ›5 = -2/7 (5)^{1/3}

Since, 5 is not a perfect cube.

Hence it is an irrational number.

**(v) **(2 – âˆš3) (2 + âˆš3)

Let us simplify,

By using the formula,

(a + b) (a – b) = (a)^{2} (b)^{2}

(2 – âˆš3) (2 + âˆš3) = (2)^{2} â€“ (âˆš3)^{2}

= 4 â€“ 3

= 1

Hence, it is a rational number.

**(vi) **(3 + âˆš5)^{2}** **

Let us simplify,

By using (a + b)^{2} = a^{2} + b^{2} + 2ab

(3 + âˆš5)^{2} = 3^{2} + (âˆš5)^{2} + 2.3.âˆš5

= 9 + 5 + 6âˆš5

= 14 + 6âˆš5

Hence, it is an irrational number.

**(vii) **(2/5 âˆš7)^{2}

Let us simplify,

(2/5 âˆš7)^{2} = (2/5 âˆš7) Ã— (2/5 âˆš7)

= 4/ 25 Ã— 7

= 28/25

Hence it is a rational number.

**(viii) **(3 – âˆš6)^{2}

Let us simplify,

By using (a – b)^{2} = a^{2} + b^{2} – 2ab

(3 – âˆš6)^{2} = 3^{2} + (âˆš6)^{2} â€“ 2.3.âˆš6

= 9 + 6 – 6âˆš6

= 15 – 6âˆš6

Hence it is an irrational number.

**8. Prove the following are irrational numbers.**

**(i) âˆ›2**

**(ii) âˆ›3**

**(iii) âˆœ5**

**Solution:**

**(i) âˆ›2**

We know that âˆ›2 = 2^{1/3}

Let us consider 2^{1/3} = p/q, where p, q are integers, q>0.

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

So,

2^{1/3} = p/q

2 = p^{3}/q^{3}

p^{3} = 2q^{3} â€¦.. (1)

We know that, 2 divides 2q^{3} then 2 divides p^{3}

So, 2 divides p

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

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

p^{3} = 2q^{3}

(2k)^{3} = 2q^{3}

8k^{3} = 2q^{3}

4k^{3} = q^{3}

We know that, 2 divides 4k^{3} then 2 divides q^{3}

So, 2 divides q

Thus p and q have a common factor â€˜2â€™.

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

Hence, âˆ›2 is an irrational number.

**(ii) **âˆ›3

We know that âˆ›3 = 3^{1/3}

Let us consider 3^{1/3} = p/q, where p, q are integers, q>0.

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

So,

3^{1/3} = p/q

3 = p^{3}/q^{3}

p^{3} = 3q^{3} â€¦.. (1)

We know that, 3 divides 3q^{3} then 3 divides p^{3}

So, 3 divides p

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

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

p^{3} = 3q^{3}

(3k)^{3} = 3q^{3}

9k^{3} = 3q^{3}

3k^{3} = q^{3}

We know that, 3 divides 9k^{3} then 3 divides q^{3}

So, 3 divides q

Thus p and q have a common factor â€˜3â€™.

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

Hence, âˆ›3 is an irrational number.

**(iii) **âˆœ5

We know that âˆœ5 = 5^{1/4}

Let us consider 5^{1/4} = p/q, where p, q are integers, q>0.

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

So,

5^{1/4} = p/q

5 = p^{4}/q^{4}

P^{4} = 5q^{4} â€¦.. (1)

We know that, 5 divides 5q^{4} then 5 divides p^{4}

So, 5 divides p

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

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

P^{4} = 5q^{4}

(5k)^{4} = 5q^{4}

625k^{4} = 5q^{4}

125k^{4} = q^{4}

We know that, 5 divides 125k^{4} then 5 divides q^{4}

So, 5 divides q

Thus p and q have a common factor â€˜5â€™.

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

Hence, âˆœ5 is an irrational number.

**9. Find the greatest and the smallest real numbers.**

**(i) 2âˆš3, 3/âˆš2, -âˆš7, âˆš15**

**(ii) -3âˆš2, 9/âˆš5, -4, 4/3 âˆš5, 3/2âˆš3**

**Solution:**

**(i) **2âˆš3, 3/âˆš2, -âˆš7, âˆš15

Let us simplify each fraction

2âˆš3 = âˆš(4Ã—3) = âˆš12

3/âˆš2 = (3Ã—âˆš2)/(âˆš2Ã—âˆš2) = 3âˆš2/2 = âˆš((9/4)Ã—2) = âˆš(9/2) = âˆš4.5

-âˆš7 = -âˆš7

âˆš15 = âˆš15

So,

The greatest real number = âˆš15

Smallest real number = -âˆš7

**(ii) **-3âˆš2, 9/âˆš5, -4, 4/3 âˆš5, 3/2âˆš3

Let us simplify each fraction

-3âˆš2 = -âˆš(9Ã—2) = -âˆš18

9/âˆš5 = (9Ã—âˆš5)/(âˆš5Ã—âˆš5) = 9âˆš5/5 = âˆš((81/25)Ã—5) = âˆš(81/5) = âˆš16.2

-4 = -âˆš16

4/3 âˆš5 = âˆš((16/9)Ã—5) = âˆš(80/9) = âˆš8.88 = âˆš8.8

3/2âˆš3 = âˆš((9/4)Ã—3) = âˆš(27/4) = âˆš6.25

So,

The greatest real number = 9âˆš5

Smallest real number = -3âˆš2

**10. Write in ascending order.**

**(i) 3âˆš2, 2âˆš3, âˆš15, 4**

**(ii) 3âˆš2, 2âˆš8, 4, âˆš50, 4âˆš3**

**Solution:**

**(i) **3âˆš2, 2âˆš3, âˆš15, 4

3âˆš2 = âˆš(9Ã—2) =âˆš18

2âˆš3 = âˆš(4Ã—3) =âˆš12

âˆš15 = âˆš15

4 = âˆš16

Now, let us arrange in ascending order

âˆš12, âˆš15, âˆš16, âˆš18

So,

2âˆš3, âˆš15, 4, 3âˆš2

**(ii) **3âˆš2, 2âˆš8, 4, âˆš50, 4âˆš3

3âˆš2 = âˆš(9Ã—2) =âˆš18

2âˆš8 = âˆš(4Ã—8) =âˆš32

4 = âˆš16

âˆš50 = âˆš50

4âˆš3 =âˆš(16Ã—3) = âˆš48

Now, let us arrange in ascending order

âˆš16, âˆš18, âˆš32, âˆš48, âˆš50

So,

4, 3âˆš2, 2âˆš8, 4âˆš3, âˆš50

**11. Write in descending order.**

**(i) 9/âˆš2, 3/2 âˆš5, 4âˆš3, 3âˆš(6/5)**

**(ii) 5/âˆš3, 7/3 âˆš2, -âˆš3, 3âˆš5, 2âˆš7**

**Solution:**

**(i) **9/âˆš2, 3/2 âˆš5, 4âˆš3, 3âˆš(6/5)

9/âˆš2 = (9Ã—âˆš2)/(âˆš2Ã—âˆš2) = 9âˆš2/2 = âˆš((81/4)Ã—2) = âˆš(81/2) = âˆš40.5

3/2 âˆš5 = âˆš((9/4)Ã—5) = âˆš(45/4) = âˆš11.25

4âˆš3 = âˆš(16Ã—3) = âˆš48

3âˆš(6/5) = âˆš((9Ã—6)/5) = âˆš(54/5) = âˆš10.8

Now, let us arrange in descending order

âˆš48, âˆš40.5, âˆš11.25, âˆš10.8

So,

4âˆš3, 9/âˆš2, 3/2 âˆš5, 3âˆš(6/5)

**(ii) **5/âˆš3, 7/3 âˆš2, -âˆš3, 3âˆš5, 2âˆš7

5/âˆš3 = âˆš(25/3) = âˆš8.33

7/3 âˆš2 = âˆš((49/9) Ã—2) = âˆš98/9 = âˆš10.88

-âˆš3 = -âˆš3

3âˆš5 = âˆš(9Ã—5) =âˆš45

2âˆš7 = âˆš(4Ã—7) = âˆš28

Now, let us arrange in descending order

âˆš45, âˆš28, âˆš10.88.., âˆš8.33.., -âˆš3

So,

3âˆš5, 2âˆš7, 7/3âˆš2, 5/âˆš3, -âˆš3

**12. Arrange in ascending order.**

**âˆ›2, âˆš3, ^{6}âˆš5**

**Solution:**

Here we can express the given expressions as:

âˆ›2 = 2^{1/3}

âˆš3 = 3^{1/2}

^{6}âˆš5 = 5^{1/6}

Let us make the roots common so,

2^{1/3}= 2^{(2}Ã— ^{1/2 }Ã—^{ 1/3)} = 4^{1/6}

3^{1/2} = 3^{(3}Ã—^{ 1/3 }Ã— ^{1/2)} = 27^{1/6}

5^{1/6} = 5^{1/6}

Now, let us arrange in ascending order,

4^{1/6}, 5^{1/6}, 27^{1/6}

So,

2^{1/3}, 5^{1/6}, 3^{1/2}

So,

âˆ›2, ^{6}âˆš5, âˆš3

EXERCISE 1.5

**1. Rationalize the following:**

**(i) 3/4âˆš5**

**(ii) 5âˆš7 / âˆš3**

**(iii) 3/(4 – âˆš7)**

**(iv) 17/(3âˆš2 + 1)**

**(v) 16/ (âˆš41 – 5)**

**(vi) 1/ (âˆš7 – âˆš6)**

**(vii) 1/ (âˆš5 + âˆš2)**

**(viii) (âˆš2 + âˆš3) / (âˆš2 – âˆš3)**

**Solution:**

**(i) **3/4âˆš5

Let us rationalize,

3/4âˆš5 = (3Ã—âˆš5) /(4âˆš5Ã—âˆš5)

= (3âˆš5) / (4Ã—5)

= (3âˆš5) / 20

**(ii) **5âˆš7 / âˆš3

Let us rationalize,

5âˆš7 / âˆš3** **= (5âˆš7Ã—âˆš3) / (âˆš3Ã—âˆš3)

= 5âˆš21/3

**(iii) **3/(4 – âˆš7)

Let us rationalize,

3/(4 – âˆš7) = [3Ã—(4 + âˆš7)] / [(4 – âˆš7) Ã— (4 + âˆš7)]

= 3(4 + âˆš7) / [4^{2} â€“ (âˆš7)^{2}]

= 3(4 + âˆš7) / [16 – 7]

= 3(4 + âˆš7) / 9

= (4 + âˆš7) / 3

**(iv) **17/(3âˆš2 + 1)

Let us rationalize,

17/(3âˆš2 + 1) = 17(3âˆš2 – 1) / [(3âˆš2 + 1) (3âˆš2 – 1)]

= 17(3âˆš2 – 1) / [(3âˆš2)^{2} – 1^{2}]

= 17(3âˆš2 – 1) / [9.2 – 1]

= 17(3âˆš2 – 1) / [18 – 1]

= 17(3âˆš2 – 1) / 17

= (3âˆš2 – 1)

**(v) **16/ (âˆš41 – 5)

Let us rationalize,

16/ (âˆš41 – 5) = 16(âˆš41 + 5) / [(âˆš41 – 5) (âˆš41 + 5)]

= 16(âˆš41 + 5) / [(âˆš41)^{2} – 5^{2}]

= 16(âˆš41 + 5) / [41 – 25]

= 16(âˆš41 + 5) / [16]

= (âˆš41 + 5)

**(vi) **1/ (âˆš7 – âˆš6)

Let us rationalize,

1/ (âˆš7 – âˆš6) = 1(âˆš7 + âˆš6) / [(âˆš7 – âˆš6) (âˆš7 + âˆš6)]

= (âˆš7 + âˆš6) / [(âˆš7)^{2} â€“ (âˆš6)^{2}]

= (âˆš7 + âˆš6) / [7 – 6]

= (âˆš7 + âˆš6) / 1

= (âˆš7 + âˆš6)

**(vii) **1/ (âˆš5 + âˆš2)

Let us rationalize,

1/ (âˆš5 + âˆš2) = 1(âˆš5 – âˆš2) / [(âˆš5 + âˆš2) (âˆš5 – âˆš2)]

= (âˆš5 – âˆš2) / [(âˆš5)^{2} â€“ (âˆš2)^{2}]

= (âˆš5 – âˆš2) / [5 – 2]

= (âˆš5 – âˆš2) / [3]

= (âˆš5 – âˆš2) /3

**(viii) **(âˆš2 + âˆš3) / (âˆš2 – âˆš3)

Let us rationalize,

(âˆš2 + âˆš3) / (âˆš2 – âˆš3) = [(âˆš2 + âˆš3) (âˆš2 + âˆš3)] / [(âˆš2 – âˆš3) (âˆš2 + âˆš3)]

= [(âˆš2 + âˆš3)^{2}] / [(âˆš2)^{2} â€“ (âˆš3)^{2}]

= [2 + 3 + 2âˆš2âˆš3] / [2 – 3]

= [5 + 2âˆš6] / -1

= – (5 + 2âˆš6)

**2. Simplify:**

**(i) (7 + 3âˆš5) / (7 – 3âˆš5)**

**(ii) (3 – 2âˆš2) / (3 + 2âˆš2)**

**(iii) (5 – 3âˆš14) / (7 + 2âˆš14)**

**Solution:**

**(i) **(7 + 3âˆš5) / (7 – 3âˆš5)

Let us rationalize the denominator, we get

(7 + 3âˆš5) / (7 – 3âˆš5) = [(7 + 3âˆš5) (7 + 3âˆš5)] / [(7 – 3âˆš5) (7 + 3âˆš5)]

** **= [(7 + 3âˆš5)^{2}] / [7^{2} â€“ (3âˆš5)^{2}]

= [7^{2} + (3âˆš5)^{2} + 2.7. 3âˆš5] / [49 â€“ 9.5]

= [49 + 9.5 + 42âˆš5] / [49 – 45]

= [49 + 45 + 42âˆš5] / [4]

= [94 + 42âˆš5] / 4

= 2[47 + 21âˆš5]/4

= [47 + 21âˆš5]/2

**(ii) **(3 – 2âˆš2) / (3 + 2âˆš2)

Let us rationalize the denominator, we get

(3 – 2âˆš2) / (3 + 2âˆš2) = [(3 – 2âˆš2) (3 – 2âˆš2)] / [(3 + 2âˆš2) (3 – 2âˆš2)]

= [(3 – 2âˆš2)^{2}] / [3^{2} â€“ (2âˆš2)^{2}]

= [3^{2} + (2âˆš2)^{2} â€“ 2.3.2âˆš2] / [9 â€“ 4.2]

= [9 + 4.2 – 12âˆš2] / [9 – 8]

= [9 + 8 – 12âˆš2] / 1

= 17 – 12âˆš2

**(iii) **(5 – 3âˆš14) / (7 + 2âˆš14)

Let us rationalize the denominator, we get

(5 – 3âˆš14) / (7 + 2âˆš14) = [(5 – 3âˆš14) (7 – 2âˆš14)] / [(7 + 2âˆš14) (7 – 2âˆš14)]

= [5(7 – 2âˆš14) â€“ 3âˆš14 (7 – 2âˆš14)] / [7^{2} â€“ (2âˆš14)^{2}]

= [35 – 10âˆš14 – 21âˆš14 + 6.14] / [49 â€“ 4.14]

= [35 – 31âˆš14 + 84] / [49 – 56]

= [119 – 31âˆš14] / [-7]

= -[119 – 31âˆš14] / 7

= [31âˆš14 â€“ 119] / 7

**3. Simplify:**

**[7âˆš3 / (âˆš10 + âˆš3)] â€“ [2âˆš5 / (âˆš6 + âˆš5)] â€“ [3âˆš2 / (âˆš15 + 3âˆš2)]**

**Solution:**

Let us simplify individually,

[7âˆš3 / (âˆš10 + âˆš3)]Let us rationalize the denominator,

7âˆš3 / (âˆš10 + âˆš3) = [7âˆš3(âˆš10 – âˆš3)] / [(âˆš10 + âˆš3) (âˆš10 – âˆš3)]

= [7âˆš3.âˆš10 – 7âˆš3.âˆš3] / [(âˆš10)^{2} â€“ (âˆš3)^{2}]

= [7âˆš30 â€“ 7.3] / [10 – 3]

= 7[âˆš30 – 3] / 7

= âˆš30 – 3

Now,

[2âˆš5 / (âˆš6 + âˆš5)]Let us rationalize the denominator, we get

2âˆš5 / (âˆš6 + âˆš5) = [2âˆš5 (âˆš6 – âˆš5)] / [(âˆš6 + âˆš5) (âˆš6 – âˆš5)]

= [2âˆš5.âˆš6 – 2âˆš5.âˆš5] / [(âˆš6)^{2} â€“ (âˆš5)^{2}]

= [2âˆš30 â€“ 2.5] / [6 – 5]

= [2âˆš30 – 10] / 1

= 2âˆš30 â€“ 10

Now,

[3âˆš2 / (âˆš15 + 3âˆš2)]Let us rationalize the denominator, we get

3âˆš2 / (âˆš15 + 3âˆš2) = [3âˆš2 (âˆš15 – 3âˆš2)] / [(âˆš15 + 3âˆš2) (âˆš15 – 3âˆš2)]

= [3âˆš2.âˆš15 – 3âˆš2.3âˆš2] / [(âˆš15)^{2} â€“ (3âˆš2)^{2}]

= [3âˆš30 â€“ 9.2] / [15 â€“ 9.2]

= [3âˆš30 – 18] / [15 – 18]

= 3[âˆš30 – 6] / [-3]

= [âˆš30 – 6] / -1

= 6 – âˆš30

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

[7âˆš3 / (âˆš10 + âˆš3)] â€“ [2âˆš5 / (âˆš6 + âˆš5)] â€“ [3âˆš2 / (âˆš15 + 3âˆš2)]= (âˆš30 â€“ 3) â€“ (2âˆš30 â€“ 10) â€“ (6 – âˆš30)

= âˆš30 â€“ 3 – 2âˆš30 + 10 â€“ 6 + âˆš30

= 2âˆš30 – 2âˆš30 â€“ 3 + 10 â€“ 6

= 1

**4. Simplify:**

**[1/(âˆš4 + âˆš5)] + [1/(âˆš5 + âˆš6)] + [1/(âˆš6 + âˆš7)] + [1/(âˆš7 + âˆš8)] + [1/(âˆš8 + âˆš9)] **

**Solution:**

Let us simplify individually,

[1/(âˆš4 + âˆš5)]Rationalize the denominator, we get

[1/(âˆš4 + âˆš5)] = [1(âˆš4 – âˆš5)] / [(âˆš4 + âˆš5) (âˆš4 – âˆš5)]= [(âˆš4 – âˆš5)] / [(âˆš4)^{2} â€“ (âˆš5)^{2}]

= [(âˆš4 – âˆš5)] / [4 – 5]

= [(âˆš4 – âˆš5)] / -1

= -(âˆš4 – âˆš5)

Now,

[1/(âˆš5 + âˆš6)]Rationalize the denominator, we get

[1/(âˆš5 + âˆš6)] = [1(âˆš5 – âˆš6)] / [(âˆš5 + âˆš6) (âˆš5 – âˆš6)]= [(âˆš5 – âˆš6)] / [(âˆš5)^{2} â€“ (âˆš6)^{2}]

= [(âˆš5 – âˆš6)] / [5 – 6]

= [(âˆš5 – âˆš6)] / -1

= -(âˆš5 – âˆš6)

Now,

[1/(âˆš6 + âˆš7)]Rationalize the denominator, we get

[1/(âˆš6 + âˆš7)] = [1(âˆš6 – âˆš7)] / [(âˆš6 + âˆš7) (âˆš6 – âˆš7)]= [(âˆš6 – âˆš7)] / [(âˆš6)^{2} â€“ (âˆš7)^{2}]

= [(âˆš6 – âˆš7)] / [6 – 7]

= [(âˆš6 – âˆš7)] / -1

= -(âˆš6 – âˆš7)

Now,

[1/(âˆš7 + âˆš8)]Rationalize the denominator, we get

[1/(âˆš7 + âˆš8)] = [1(âˆš7 – âˆš8)] / [(âˆš7 + âˆš8) (âˆš7 – âˆš8)]= [(âˆš7 – âˆš8)] / [(âˆš7)^{2} â€“ (âˆš8)^{2}]

= [(âˆš7 – âˆš8)] / [7 – 8]

= [(âˆš7 – âˆš8)] / -1

= -(âˆš7 – âˆš8)

Now,

[1/(âˆš8 + âˆš9)]Rationalize the denominator, we get

[1/(âˆš8 + âˆš9)] = [1(âˆš8 – âˆš9)] / [(âˆš8 + âˆš9) (âˆš8 – âˆš9)]= [(âˆš8 – âˆš9)] / [(âˆš8)^{2} â€“ (âˆš9)^{2}]

= [(âˆš8 – âˆš9)] / [8 – 9]

= [(âˆš8 – âˆš9)] / -1

= -(âˆš8 – âˆš9)

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

[1/(âˆš4 + âˆš5)] + [1/(âˆš5 + âˆš6)] + [1/(âˆš6 + âˆš7)] + [1/(âˆš7 + âˆš8)] + [1/(âˆš8 + âˆš9)]= -(âˆš4 – âˆš5) + -(âˆš5 – âˆš6) + -(âˆš6 – âˆš7) + -(âˆš7 – âˆš8) + -(âˆš8 – âˆš9)

= -âˆš4 + âˆš5 – âˆš5 + âˆš6 – âˆš6 + âˆš7 – âˆš7 + âˆš8 – âˆš8 + âˆš9

= -âˆš4 + âˆš9

= -2 + 3

= 1

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

**(i) [3 – **âˆš5**] / [3 + 2**âˆš5**] = -19/11 + a**âˆš5

**(ii) [âˆš2 + âˆš3] / [3âˆš2 – 2âˆš3] = a – bâˆš6**

**(iii) {[7 + âˆš5]/[7 – âˆš5]} â€“ {[7 – âˆš5]/[7 + âˆš5]} = a + 7/11 bâˆš5**

**Solution:**

**(i) **[3 – âˆš5] / [3 + 2âˆš5] = -19/11 + aâˆš5

Let us consider LHS

[3 – âˆš5] / [3 + 2âˆš5]Rationalize the denominator,

[3 – âˆš5] / [3 + 2âˆš5] = [(3 – âˆš5) (3 – 2âˆš5)] / [(3 + 2âˆš5) (3 – 2âˆš5)]= [3(3 – 2âˆš5) â€“ âˆš5(3 – 2âˆš5)] / [3^{2} â€“ (2âˆš5)^{2}]

= [9 – 6âˆš5 – 3âˆš5 + 2.5] / [9 â€“ 4.5]

= [9 – 6âˆš5 – 3âˆš5 + 10] / [9 â€“ 20]

= [19 – 9âˆš5] / -11

= -19/11 + 9âˆš5/11

So when comparing with RHS

-19/11 + 9âˆš5/11 = -19/11 + aâˆš5

Hence, value of a = 9/11

**(ii) **[âˆš2 + âˆš3] / [3âˆš2 – 2âˆš3] = a – bâˆš6

Let us consider LHS

[âˆš2 + âˆš3] / [3âˆš2 – 2âˆš3]Rationalize the denominator,

[âˆš2 + âˆš3] / [3âˆš2 – 2âˆš3] = [(âˆš2 + âˆš3) (3âˆš2 + 2âˆš3)] / [(3âˆš2 – 2âˆš3) (3âˆš2 + 2âˆš3)]= [âˆš2(3âˆš2 + 2âˆš3) + âˆš3(3âˆš2 + 2âˆš3)] / [(3âˆš2)^{2} â€“ (2âˆš3)^{2}]

= [3.2 + 2âˆš2âˆš3 + 3âˆš2âˆš3 + 2.3] / [9.2 â€“ 4.3]

= [6 + 2âˆš6 + 3âˆš6 + 6] / [18 – 12]

= [12 + 5âˆš6] / 6

= 12/6 + 5âˆš6/6

= 2 + 5âˆš6/6

= 2 â€“ (-5âˆš6/6)

So when comparing with RHS

2 â€“ (-5âˆš6/6) = a – bâˆš6

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

**(iii) **{[7 + âˆš5]/[7 – âˆš5]} â€“ {[7 – âˆš5]/[7 + âˆš5]} = a + 7/11 bâˆš5

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + âˆš5]/[7 – âˆš5]}

Rationalize the denominator,

[7 + âˆš5]/[7 – âˆš5] = [(7 + âˆš5) (7 + âˆš5)] / [(7 – âˆš5) (7 + âˆš5)]= [(7 + âˆš5)^{2}] / [7^{2} â€“ (âˆš5)^{2}]

= [7^{2} + (âˆš5)^{2} + 2.7.âˆš5] / [49 – 5]

= [49 + 5 + 14âˆš5] / [44]

= [54 + 14âˆš5] / 44

Now,

{[7 – âˆš5]/[7 + âˆš5]}

Rationalize the denominator,

[7 – âˆš5]/[7 + âˆš5] = (7 – âˆš5) (7 – âˆš5)] / [(7 + âˆš5) (7 – âˆš5)]= [(7 – âˆš5)^{2}] / [7^{2} â€“ (âˆš5)^{2}]

= [7^{2} + (âˆš5)^{2} – 2.7.âˆš5] / [49 – 5]

= [49 + 5 – 14âˆš5] / [44]

= [54 – 14âˆš5] / 44

So, according to the question

{[7 + âˆš5]/[7 – âˆš5]} â€“ {[7 – âˆš5]/[7 + âˆš5]}

By substituting the obtained values,

= {[54 + 14âˆš5] / 44} â€“ {[54 – 14âˆš5] / 44}

= [54 + 14âˆš5 â€“ 54 + 14âˆš5]/44

= 28âˆš5/44

= 7âˆš5/11

So when comparing with RHS

7âˆš5/11 = a + 7/11 bâˆš5

Hence, value of a = 0 and b = 1

**6. Simplify:**

**{[7 + 3âˆš5] / [3 + âˆš5]} â€“ {[7 – 3âˆš5] / [3 – âˆš5]} = p + qâˆš5**

**Solution:**

Let us consider LHS

Since there are two terms, let us solve individually

{[7 + 3âˆš5] / [3 + âˆš5]}

Rationalize the denominator,

[7 + 3âˆš5] / [3 + âˆš5] = [(7 + 3âˆš5) (3 – âˆš5)] / [(3 + âˆš5) (3 – âˆš5)]= [7(3 – âˆš5) + 3âˆš5(3 – âˆš5)] / [3^{2} â€“ (âˆš5)^{2}]

= [21 – 7âˆš5 + 9âˆš5 â€“ 3.5] / [9 – 5]

= [21 + 2âˆš5 – 15] / [4]

= [6 + 2âˆš5] / 4

= 2[3 + âˆš5]/4

= [3 + âˆš5] /2

Now,

{[7 – 3âˆš5] / [3 – âˆš5]}

Rationalize the denominator,

[7 – 3âˆš5] / [3 – âˆš5] = [(7 – 3âˆš5) (3 + âˆš5)] / [(3 – âˆš5) (3 + âˆš5)]= [7(3 + âˆš5) â€“ 3âˆš5(3 + âˆš5)] / [3^{2} â€“ (âˆš5)^{2}]

= [21 + 7âˆš5 – 9âˆš5 â€“ 3.5] / [9 – 5]

= [21 – 2âˆš5 – 15] / 4

= [6 – 2âˆš5]/4

= 2[3 – âˆš5]/4

= [3 – âˆš5]/2

So, according to the question

{[7 + 3âˆš5] / [3 + âˆš5]} â€“ {[7 – 3âˆš5] / [3 – âˆš5]}

By substituting the obtained values,

= {[3 + âˆš5] /2} â€“ {[3 – âˆš5] /2}

= [3 + âˆš5 â€“ 3 + âˆš5]/2

= [2âˆš5]/2

= âˆš5

So when comparing with RHS

âˆš5 = p + qâˆš5

Hence, value of p = 0 and q = 1

**7. If âˆš2 = 1.414, âˆš3 = 1.732, find**

**(i) âˆš2/(2 + âˆš2)**

**(ii) 1/(âˆš3 + âˆš2)**

**Solution:**

**(i) **âˆš2/(2 + âˆš2)

By rationalizing the denominator,

âˆš2/(2 + âˆš2)** **= [âˆš2(2 – âˆš2)] / [(2 + âˆš2) (2 – âˆš2)]

= [2âˆš2 â€“ 2] / [2^{2} â€“ (âˆš2)^{2}]

= [2âˆš2 – 2] / [4 – 2]

= 2[âˆš2 – 1] / 2

= âˆš2 â€“ 1

= 1.414 â€“ 1

= 0.414

**(ii) **1/(âˆš3 + âˆš2)

By rationalizing the denominator,

1/(âˆš3 + âˆš2) = [1(âˆš3 – âˆš2)] / [(âˆš3 + âˆš2) (âˆš3 – âˆš2)]

= [(âˆš3 – âˆš2)] / [(âˆš3)^{2} â€“ (âˆš2)^{2}]

= [(âˆš3 – âˆš2)] / [3 – 2]

= [(âˆš3 – âˆš2)] / 1

= (âˆš3 – âˆš2)

= 1.732 â€“ 1.414

= 0.318

**8. If a = 2 + âˆš3, find 1/a, (a â€“ 1/a)**

**Solution:**

Given:

a = 2 + âˆš3

So,

1/a = 1/ (2 + âˆš3)

By rationalizing the denominator,

1/ (2 + âˆš3) = [1(2 – âˆš3)] / [(2 + âˆš3) (2 – âˆš3)]

= [(2 – âˆš3)] / [2^{2} â€“ (âˆš3)^{2}]

= [(2 – âˆš3)] / [4 – 3]

= (2 – âˆš3)

Then,

a â€“ 1/a = 2 + âˆš3 â€“ (2 – âˆš3)

= 2 + âˆš3 â€“ 2 + âˆš3

= 2âˆš3

**9. Solve:**

**If x = 1 – âˆš2, find 1/x, (x â€“ 1/x) ^{4}**

**Solution:**

Given:

x = 1 – âˆš2

so,

1/x = 1/(1 – âˆš2)

By rationalizing the denominator,

1/ (1 – âˆš2) = [1(1 + âˆš2)] / [(1 – âˆš2) (1 + âˆš2)]

= [(1 + âˆš2)] / [1^{2} â€“ (âˆš2)^{2}]

= [(1 + âˆš2)] / [1 – 2]

= (1 + âˆš2) / -1

= -(1 + âˆš2 )

Then,

(x â€“ 1/x)^{4} = [1 – âˆš2 â€“ (-1 – âˆš2)]^{4}

= [1 – âˆš2 + 1 + âˆš2]^{4}

= 2^{4}

= 16

**10. Solve: **

**If x = 5 – 2âˆš6, find 1/x, (x ^{2} â€“ 1/x^{2}) **

**Solution:**

Given:

x = 5 â€“ 2âˆš6

so,

1/x = 1/(5 – 2âˆš6)

By rationalizing the denominator,

1/(5 – 2âˆš6) = [1(5 + 2âˆš6)] / [(5 – 2âˆš6) (5 + 2âˆš6)]

= [(5 + 2âˆš6)] / [5^{2} â€“ (2âˆš6)^{2}]

= [(5 + 2âˆš6)] / [25 â€“ 4.6]

= [(5 + 2âˆš6)] / [25 – 24]

= (5 + 2âˆš6)

Then,

x + 1/x = 5 â€“ 2âˆš6 + (5 + 2âˆš6)

= 10

Square on both sides we get

(x + 1/x)^{2} = 10^{2}

x^{2} + 1/x^{2} + 2x.1/x = 100

x^{2} + 1/x^{2} + 2 = 100

x^{2} + 1/x^{2} = 100 â€“ 2

= 98

**11. If p = (2-âˆš5)/(2+âˆš5) and q = (2+âˆš5)/(2-âˆš5), find the values of**

**(i) p + q**

**(ii) p â€“ q **

**(iii) p ^{2} + q^{2}**

**(iv) p ^{2} â€“ q^{2}**

**Solution:**

Given:

p = (2-âˆš5)/(2+âˆš5) and q = (2+âˆš5)/(2-âˆš5)

**(i) **p + q

So by rationalizing the denominator, we get

= [(2 – âˆš5)^{2} + (2 + âˆš5)^{2}] / [2^{2} â€“ (âˆš5)^{2}]

= [4 + 5 – 4âˆš5 + 4 + 5 + 4âˆš5] / [4 – 5]

= [18]/-1

= -18

**(ii) **p â€“ q

So by rationalizing the denominator, we get

= [(2 – âˆš5)^{2} – (2 + âˆš5)^{2}] / [2^{2} â€“ (âˆš5)^{2}]

= [4 + 5 – 4âˆš5 â€“ (4 + 5 + 4âˆš5)] / [4 – 5]

= [9 – 4âˆš5 â€“ 9 – 4âˆš5] / -1

= [-8âˆš5]/-1

= 8âˆš5

**(iii) **p^{2} + q^{2}

We know that (p + q)^{2} = p^{2} + q^{2} + 2pq

So,

p^{2} + q^{2} = (p + q)^{2} â€“ 2pq

pq = [(2-âˆš5)/(2+âˆš5)] Ã— [(2+âˆš5)/(2-âˆš5)]

= 1

p + q = -18

so,

p^{2} + q^{2} = (p + q)^{2} â€“ 2pq

= (-18)^{2} â€“ 2(1)

= 324 â€“ 2

= 322

**(iv) **p^{2} â€“ q^{2}

We know that, p^{2} â€“ q^{2} = (p + q) (p – q)

So, by substituting the values

p^{2} â€“ q^{2} = (p + q) (p – q)

= (-18) (8âˆš5)

= -144âˆš5

**12. If x = (âˆš2 – 1)/( âˆš2 + 1) and y = (âˆš2 + 1)/( âˆš2 – 1), find **

**(i) x + y**

**(ii) xy**

**Solution:**

Given:

x = (âˆš2 – 1)/( âˆš2 + 1) and y = (âˆš2 + 1)/( âˆš2 – 1)

**(i) **x + y

= [(âˆš2 – 1)/( âˆš2 + 1)] + [(âˆš2 + 1)/( âˆš2 – 1)]

By rationalizing the denominator,

= [(âˆš2 – 1)^{2} + (âˆš2 + 1)^{2}] / [(âˆš2)^{2} – 1^{2}]

= [2 + 1 – 2âˆš2 + 2 + 1 + 2âˆš2] / [2 – 1]

= [6] / 1

= 6

**(ii) **xy

= 1