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

**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

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

= 7/24

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

By taking LCM and rationalizing, we get

= 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

= -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

= -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 × 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 = p^{2} / q^{2}

p^{2} = 5q^{2} **…. (1)**

As we know, ‘5’ divides 5q^{2}, so ‘5’ divides p^{2} 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

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

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

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

As we know, ‘5’ divides 5k^{2}, so ‘5’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 7q^{2} **…. (1)**

As we know, ‘7’ divides 7q^{2}, so ‘7’ divides p^{2} 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

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

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

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

As we know, ‘7’ divides 7k^{2}, so ‘7’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 6q^{2} **…. (1)**

As we know, ‘2’ divides 6q^{2}, so ‘2’ divides p^{2} 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

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

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

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

As we know, ‘2’ divides 2k^{2}, so ‘2’ divides 3q^{2} as well.

‘2’ should either divide 3 or divide q^{2}.

But ‘2’ does not divide 3. ‘2’ divides q^{2} 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 = p^{2} / q^{2}

q^{2} = 11p^{2} **…. (1)**

As we know, ‘11’ divides 11p^{2}, so ‘11’ divides q^{2} 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

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

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

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

As we know, ‘11’ divides 11k^{2}, so ‘11’ divides p^{2} 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 = p^{2} / q^{2}

p^{2} = 2q^{2} **…. (1)**

As we know, ‘2’ divides 2q^{2}, so ‘2’ divides p^{2} 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

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

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

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

As we know, ‘2’ divides 2k^{2}, so ‘2’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 3q^{2} **…. (1)**

As we know, ‘3’ divides 3q^{2}, so ‘3’ divides p^{2} 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

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

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

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

As we know, ‘3’ divides 3k^{2}, so ‘3’ divides q^{2} 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 = p^{2} / q^{2}

p^{2} = 5q^{2} **…. (1)**

As we know, ‘5’ divides 5q^{2}, so ‘5’ divides p^{2} 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

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

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

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

As we know, ‘5’ divides 5k^{2}, so ‘5’ divides q^{2} 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 = r^{2} + (√2)^{2} – 2r√2

5 = r^{2} + 2 – 2√2r

5 – 2 = r^{2} – 2√2r

r^{2} – 3 = 2√2r

(r^{2} – 3)/2r = √2

We know that, ‘r’ is rational, ‘(r^{2} – 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} + 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 = √10cm

√17 = √(16 + 1) = √((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 = √17cm

**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 = (4×8 + 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 × 5 × 5 × 5 × 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 × 2 × 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 × 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

625 = 5 × 5 × 5 × 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 **÷ **7) / (210 **÷ **7)

= 11/30

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: 2^{n}, 5^{n}

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) 2²×7/5 ^{4}**

**(vi) 237/1500**

**Solution**

**(i) **17/8

Denominator, 8 = 2 × 2 × 2 = 2^{3}

It is a terminating decimal.

When we divide 17/8, we get

17/8 = 2.125

**(ii) **13/3125

3125 = 5 × 5 × 5 × 5 × 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 × 2 × 2 × 2 × 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 **÷ **3) / (15 **÷ **3)

= 2/5

Since the denominator is 5,

It is terminating decimal.

6/15 = 0.4

**(v) (**2²×7)/5^{4}

We know that the denominator is 5^{4}

It is a terminating decimal.

**(**2²×7)/5^{4} = (2 × 2 × 7) / (5 × 5 × 5 × 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 **÷ **3) / (1500 **÷ **3)

= 79/500

Since the denominator is 500,

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

= 2^{2} × 5^{3}

It is terminating decimal.

237/1500= 79/500 = 0.1518

**6. Write the denominator of the rational number 257/5000 in the form 2 ^{m }× 5^{n }where 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:

5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5

= 2^{3} × 5^{4}

257/5000 = 257/(2^{3} × 5^{4})

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

1/7 = 0.142857142857

Since it is recurring,

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

**Solution**

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

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

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

Let x == 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

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

Let x == 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

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

**(i) √23**

**(ii) √225**

**(iii) 0.3796**

**(iv) 7.478478**

**(v) 1.101001000100001…**

**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.

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.

**10. The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form p/q, where p, q are integers, q≠ 0 and p, q are co-prime, then what can you say about the prime factors of q?**

**(i) 37.09158**

**(iii) 8.9010010001…**

**(iv) 2.3476817681…**

**Solution**

(i) 37.09158

We know that

It has terminating decimal

Here

It is a rational number and factors of q will be 2 or 5 or both.

We know that

It has non-terminating recurring decimals

Here

It is a rational number.

(iii) 8.9010010001…

We know that

It has non-terminating, non-recurring decimal.

Here

It is not a rational number.

(iv) 2.3476817681…

We know that

It has non-terminating, recurring decimal.

Here

It is a rational number and the factors of q are prime factors other than 2 and 5.

**11. Insert an irrational number between the following.**

**(i) 1/3 and ½**

**(ii) -2/5 and ½**

**(iii) 0 and 0.1**

**Solution**

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

1/3 = 0.333…

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

-2/5 = -0.4

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…

**12. 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

**13. 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…

**14. 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.

**15. Find two rational numbers between 2√3 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 √15 are [√12.25, √12.96] or [√3.5, √3.6].

**16. Insert irrational numbers between √5 and √7.**

**Solution**

Since, 5 < 6 < 7

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

**17. 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} = a^{2} + b^{2} + 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} = a^{2} + b^{2} – 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 + 2^{1/3}

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

(3 + √5)^{2} = 3^{2} + (√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} = a^{2} + b^{2} – 2ab

(3 – √6)^{2} = 3^{2} + (√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 = 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} **… (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 ‘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 = 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} ….. (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 ‘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 = 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}** ... (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 ‘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 = 2^{1/3}

√3 = 3^{1/2}

^{6}√5 = 5^{1/6}

Let us make the roots common so,

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

3^{1/2} = 3^{(3}×^{ 1/3 }× ^{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,

∛2< ^{6}√5< √3

**Exercise 1.5**

**1. Rationalize the denominator of 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) / [4^{2} – (√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} – 1^{2}]

= 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} – 5^{2}]

= 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 each of the following by rationalizing the denominator:**

**(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}] / [7^{2} – (3√5)^{2}]

= [7^{2} + (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}] / [3^{2} – (2√2)^{2}]

= [3^{2} + (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)] / [7^{2} – (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. Give a and b are rational numbers. Find 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)] / [3^{2} – (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}] / [7^{2} – (√5)^{2}]

= [7^{2} + (√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}] / [7^{2} – (√5)^{2}]

= [7^{2} + (√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. If {[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]} = p + q√5, find the value of p and q where p and q are rational numbers.**

**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)] / [3^{2} – (√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)] / [3^{2} – (√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. Rationalise the denominator of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732, upto three places of decimal:**

**(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] / [2^{2} – (√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)] / [2^{2} – (√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)] / [1^{2} – (√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}

= 2^{4}

= 16

**10. Solve:**

**If x = 5 – 2√6, find 1/x, (x ^{2} – 1/x^{2})**

**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)] / [5^{2} – (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} = 10^{2}

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

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

x^{2} + 1/x^{2} = 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) p ^{2} + q^{2}**

**(iv) p ^{2} – q^{2}**

**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}] / [2^{2} – (√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}] / [2^{2} – (√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) **p^{2} + q^{2}

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

So,

p^{2} + q^{2} = (p + q)^{2} – 2pq

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

= 1

p + q = -18

so,

p^{2} + q^{2} = (p + q)^{2} – 2pq

= (-18)^{2} – 2(1)

= 324 – 2

= 322

**(iv) **p^{2} – q^{2}

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

So, by substituting the values

p^{2} – q^{2} = (p + q) (p – q)

= (-18) (8√5)

= -144√5

**12. If x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1), find the value of x ^{2} + 5xy + y^{2}.**

**Solution**

Given:

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

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

By rationalizing the denominator,

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

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

= [6] / 1

= 6

xy** = **[(√2 – 1)/( √2 + 1)] × [(√2 + 1)/( √2 – 1)]

= 1

We know that

x^{2} + 5xy + y^{2} = x^{2 }+ y^{2} + 2xy + 3xy

It can be written as

= (x + y)^{2} + 3xy

Substituting the values

= 6^{2} + 3 × 1

So we get

= 36 + 3

= 39

**Chapter test**

**1. Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals:**

**(i) 13/45**

**(ii) -5/56**

**(iii) 7/125**

**(iv) -23/80**

**(v) – 15/66**

**In case of terminating decimals, write their decimal expansions.**

**Solution**

**(i)** We know that

The fraction whose denominator is the multiple of 2 or 5 or both is a terminating decimal

In 13/45

45 = 3 × 3 × 5

Hence, it is not a terminating decimal.

**(ii)** In -5/56

56 = 2 × 2 × 2 × 7

Hence, it is not a terminating decimal.

**(iii)** In 7/125

125 = 5 × 5 × 5

We know that

Hence, it is a terminating decimal.

**(iv)** In -23/80

80 = 2 × 2 × 2 × 2 × 5

We know that

Hence, it is a terminating decimal.

**(v) **In – 15/66

66 = 2 × 3 × 11

Hence, it is not a terminating decimal.

**2. Express the following recurring decimals as vulgar fractions:**

**Solution**

**(i)** We know that

Now multiply both sides of equation (1) by 10

10x = 13.4545 **... (2)**

Again multiply both sides of equation (2) by 100

1000x = 1345.4545 **… (3)**

By subtracting equation (2) from (3)

990x = 1332

By further calculation

x = 1332/990 = 74/55

**(ii)** We know that

Now multiply both sides of equation (1) by 1000

1000x = 2357.357357 **... (2)**

By subtracting equation (1) from (2)

999x = 2355

By further calculation

x = 2355/999

**3. Insert a rational number between 5/9 and 7/13, and arrange in ascending order.**

**Solution**

We know that

A rational number between 5/9 and 7/13

= 64/117Here,

Therefore, in ascending order – 7/13, 64/117, 5/9.

**4. Insert four rational numbers between 4/5 and 5/6.**

**Solution**

We know that

Rational numbers between 4/5 and 5/6

Here LCM of 5, 6 = 30

So the four rational numbers are

121/150, 122/150, 123/150, 124/150

By further simplification

121/150, 61/75, 41/50, 62/75

**5. Prove that the reciprocal of an irrational number is irrational.**

**Solution**

Consider x as an irrational number

Reciprocal of x is 1/x

If 1/x is a non-zero rational number

Then x × 1/x will also be an irrational number.

We know that the product of a non-zero rational number and irrational number is also irrational.

If x × 1/x = 1 is rational number

Our assumption is wrong

So 1/x is also an irrational number.

Therefore, the reciprocal of an irrational number is also an irrational number.

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

**Solution**

**(i)** √8

If √8 is a rational number

Consider √8 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring on both sides

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

So we get

p^{2} = 8q^{2}

We know that

8p^{2} is divisible by 8

p^{2} is also divisible by 8

p is divisible by 8

Consider p = 8k where k is an integer

By squaring on both sides

p^{2} = (8k)^{2}

p^{2} = 64k^{2}

We know that

64k^{2} is divisible by 8

p^{2} is divisible by 8

p is divisible by 8

Here p and q both are divisible by 8

So our supposition is wrong

Therefore, √8 is an irrational number.

**(ii)** √14

If √14 is a rational number

Consider √14 = p/q where p and q are integers

q ≠ 0 and p and q have no common factor

By squaring on both sides

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

So we get

p^{2} = 14q^{2} **... (1)**

We know that

p^{2} is also divisible by 2

p is divisible by 2

Consider p = 2m

Substitute the value of p in equation (1)

(2m)^{2} = 13q^{2}

So we get

4m^{2} = 14q^{2}

2m^{2} = 7q^{2}

We know that

q^{2} is divisible by 2

q is divisible by 2

Here p and q have 2 as the common factor which is not possible

Therefore, √14 is an irrational number.

Consider = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By cubing on both sides

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

So we get

p^{3} = 2q^{3} **….. (1)**

We know that

2q^{3} is also divisible by 2

p^{3} is divisible by 2

p is divisible by 2

Consider p = 2k where k is an integer

By cubing both sides

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

p^{3} = 8k^{3}

So we get

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

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

We know that

4k^{3} is divisible by 2

q^{3} is divisible by 2

q is divisible by 2

Here p and q are divisible by 2

So our supposition is wrong

Therefore, is an irrational number.

**7. Prove that √3 is a rational number. Hence show that 5 – √3 is an irrational number.**

**Solution**

If √3 is a rational number

Consider √3 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring both sides

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

So we get

P^{2} = 3q^{2}

We know that

3q^{2} is divisible by 3

p^{2} is divisible by 3

p is divisible by 3

Consider p = 3 where k is an integer

By squaring on both sides

P^{2} = 9k^{2}

9k^{2} is divisible by 3

p^{2} is divisible by 3

3q^{2} is divisible by 3

q^{2} is divisible by 3

q is divisible by 3

Here p and q are divisible by 3

So our supposition is wrong

Therefore, √3 is an irrational number.

In 5 – √3

5 is a rational number

√3 is an irrational number (proved)

We know that

Difference of a rational number and irrational number is also an irrational number

So 5 – √3 is an irrational number.

Therefore, it is proved.

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

**(i) 3 + √5**

**(ii) 15 – 2√7**

**Solution**

**(i)** If 3 + √5 is a rational number say x

Consider 3 + √5 = x

It can be written as

√5 = x – 3

Here x – 3 is a rational number

√5 is also a rational number.

Consider √5 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring both sides

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

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

We know that

5q^{2} is divisible by 5

p^{2} is divisible by 5

p is divisible 5

Consider p = 5k where k is an integer

By squaring on both sides

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

So we get

5q^{2} = 25k^{2}

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

Here

5k^{2} is divisible by 5

q^{2} is divisible by 5

q is divisible by 5

Here p and q are divisible by 5

So our supposition is wrong

√5 is an irrational number

3 + √5 is also an irrational number.

Therefore, it is proved.

(ii) If 15 – 2√7 is a rational number say x

Consider 15 – 2√7 = x

It can be written as

2√7 = 15 – x

So we get

√7 = (15 – x)/ 2

Here

(15 – x)/ 2 is a rational number

√7 is a rational number

Consider √7 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring on both sides

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

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

Here

7q^{2} is divisible by 7

p^{2} is divisible by 7

p is divisible by 7

Consider p = 7k where k is an integer

By squaring on both sides

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

It can be written as

7q^{2} = 49k^{2}

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

Here

7k^{2} is divisible by 7

q^{2} is divisible by 7

q is divisible by 7

Here p and q are divisible by 7

So our supposition is wrong

√7 is an irrational number

15 – 2√7 is also an irrational number.

Therefore, it is proved.

By rationalizing the denominator

Here

¾ is a rational number and √5/4 is an irrational number

We know that

Sum of a rational and an irrational number is an irrational number.

Therefore, it is proved.

**9. Rationalise the denominator of the following:**

**Solution**

**10. If p, q are rational numbers and p – √15q = 2√3 – √5/4√3 – 3√5, find the values of p and q.**

**Solution**

It is given that

By comparing both sides

p = 3 and q = -2/3

**11. If x = 1/3 + 2√2, then find the value of x – 1/x.**

**Solution**

Here

1/x = 3 + 2√2/1 = √3 + 2√2

We know that

x – 1/x = (3 – 2√2) – (3 + 2√2)

By further calculation

= 3 – 2√2 – 3 – 2√2

So we get

= -4√2

13. (i) If x = 7+3√5/7-3√5, find the value of x^{2} + 1/x^{2}

^{2}+ xy + y

^{2}

^{2}+

**Solution**

Dividing by 2

=1000 – 30

= 970

**13. Write the following real numbers in descending order:**

**Solution**

We know that

√2 = √2

3.5 = √12.25

√10 = √10

Writing the above numbers in descending order

√18.75, √12.25, √10, √2, – √12.5

So we get

5/2 √3, 3.5, √10, √2, -5/√2

**14. Find a rational number and an irrational number between √3 and √5.**

**Solution**

Let (√3)^{2} = 3 and (√5)^{2} = 5

(i) There exists a rational number 4 which is the perfect square of a rational number 2.

(ii) There can be much more rational numbers which are perfect squares.

We know that, one irrational number between √3 and √5 = ½ (√3 + √5) = (√3 + √5)/2

**15. Insert three irrational numbers between 2√3 and 2√5, and arrange in descending order.**

**Solution**

Take the square

(2√3)^{2} = 12 and (2√5)^{2} = 20

So the number 13, 15, 18 lie between 12 and 20 between (√12)^{2} and (√20)^{2}

√13, √15, √18 lie between 2√3 and 2√5

Therefore, three irrational numbers between

2√3 and 2√5 are √13, √15, √18 or √13, √15 and 3√2.

Here

√20 ˃ √18 ˃ √15 ˃ √13 ˃ √12 or,

2√5 ˃ 3√2 ˃ √15 ˃ √13 ˃ 2√3

Therefore, the descending order: 2√5, 3√2, √15, √13 and 2√3.

**16. Give an example each of two different irrational numbers, whose**

**(i) sum is an irrational number.**

**(ii) product is an irrational number.**

**Solution**

**(i) **Consider a = √2 and b = √3 as two irrational numbers

Here,

a + b = √2 + √3 is also an irrational number.

**(ii)** Consider a = √2 and b = √3 as two irrational numbers

Here,

ab = √2 √3 = √6 is also an irrational number.

**17. Give an example of two different irrational numbers, a and b, where a/b is a rational number.**

**Solution**

Consider a = 3√2 and b = 5√2 as two different irrational numbers

Here

a/b = 3√2/5√2 = 3/2 is a rational number.

**18. If 34.0356 is expressed in the form p/q, where p and q are coprime integers, then what can you say about the factorization of q?**

**Solution**

We know that,

34.0356 = 340356/10000 (in p/q form)

= 85089/2500

Here,

85089 and 2500 are coprime integers

So the factorization of q = 2500 = 2^{2}× 5^{4}

Is of the form (2^{m} × 5^{n})

where m and n are positive or non-negative integers.

**19. In each case, state whether the following numbers are rational or irrational. If they are rational and expressed in the form p/q, where p and q are coprime integers, then what can you say about the prime factors of q?**

**(i) 279.034**

**(iii) 3.010010001…**

**(iv) 39.546782**

**(v) 2.3476817681…**

**(vi) 59.120120012000…**

**Solution**

**(i)** 279.034 is a rational number because it has terminating decimals

279.034 = 279034/1000 (in p/q form)

= 139517/500 (Dividing by 2)

We know that

Factors of 500 = 2 × 2 × 5 × 5 × 5 = 2^{2} × 5^{3}

Which is of the form 2^{m} × 5^{n} where m and n are positive integers.

It is a rational number as it has recurring or repeating decimals

= 76.17893 17893 17893 …..

100000x = 7617893.178931789317893…..

By subtraction

99999x = 7617817

x = 7617817/99999 which is of p/q form

We know that

Prime factor of 99999 = 3 × 3 × 11111

q has factors other than 2 or 5 i.e. 3^{2} × 11111

**(iii)** 3.010010001….

It is neither terminating decimal nor repeating

Therefore, it is an irrational number.

**(iv)** 39.546782

It is terminating decimal and is a rational number

39.546782 = 39546782/1000000 (in p/q form)

= 19773391/500000

We know that p and q are coprime

Prime factors of q = 2^{5} × 5^{6}

Is of the form 2^{m} × 5^{n} where m and n are positive integers

**(v)** 2.3476817681…

Is neither terminating nor repeated decimal

Therefore, it is an irrational number.

**(vi)** 59.120120012000….

It is neither terminating decimal nor repeated

Therefore, it is an irrational number.