# Class 9 RD Sharma Solutions – Chapter 1 Number System – Exercise 1.4

• Last Updated : 04 Mar, 2021

### Question 1: Define an irrational number.

Solution:

A real number that cannot be expressed in the form of fractions i.e. p/q, where p and q are integers and q ≠ 0. It is a non-terminating or non-repeating decimal. i.e. for example:

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1.1000120010211…..

### Question 2: Explain, how irrational numbers differ from rational numbers?

Solution:

An irrational number is a real number that cannot be expressed in the form of fractions i.e. p/q, where p and q are integers and q ≠ 0 i.e it cannot be expressed as a ratio of integers. It is a non-terminating or non-repeating decimal.

For example, √2 is an irrational number

A rational number is a real number that can be expressed as a fraction and as a decimal i.e. it can be expressed as a ratio of integers. It is a terminating or repeating decimal.

For examples: 0.101 and 5/4 are rational numbers

### Question 3: Examine, whether the following numbers are rational or irrational:

(i) √7

(ii) √4

(iii) 2 + √3

(iv) √3 + √2

(v) √3 + √5

(vi) (√2 – 2)2

(vii) (2 – √2)(2 + √2)

(viii) (√3 + √2)2

(ix) √5 – 2

(x) √23

(xi) √225

(xii) 0.3796

(xiii) 7.478478……

(xiv) 1.101001000100001……

Solution:

(i) √7

Given: √7

Since, it is not a perfect square root,

Therefore, it is an irrational number.

(ii) √4

Given: √4

Since, it is a perfect square root of 2.

Therefore, 2 can be expressed in the form of 2/1, thus it is a rational number.

(iii) 2 + √3

Given: 2 + √3

Here, 2 is a rational number, and √3 i is not a perfect square thus it is an irrational number.

Since, the sum of a rational and irrational number is always an irrational number.

Therefore, 2 + √3 is an irrational number.

(iv) √3 + √2

Given: √3 + √2

Here, √3 is not a perfect square thus it is an irrational number.

Similarly, √2 is not a perfect square, thus it is an irrational number.

Since, the sum of two irrational numbers is always an irrational number.

Therefore, √3 + √2 is an irrational number.

(v) √3 + √5

Given: √3 + √5

Here, √3 is not a perfect square thus it is an irrational number

Similarly, √5 is not a perfect square thus it is an irrational number.

Since, the sum of two irrational numbers is always an irrational number.

Therefore, √3 + √5 is an irrational number.

(vi) (√2 – 2)2

Given: (√2 – 2)2

(√2 – 2)2 = 2 + 4 – 4 √2

= 6 – 4 √2

Here, 6 is a rational number but 4√2 is an irrational number.

Since, the sum of a rational and irrational number is always an irrational number.

Therefore, (√2 – 2)2 is an irrational number.

(vii) (2 – √2)(2 + √2)

Given: (2 – √2)(2 + √2)

(2 – √2)(2 + √2) = ((2)2 − (√2)2) [As, (a + b)(a – b) = a2 – b2]

= 4 – 2

= 2 or 2/1

Since, 2 is a rational number,

Therefore, (2 – √2)(2 + √2) is a rational number.

(viii) (√3 + √2)2

Given: (√3 + √2)2

(√3 + √2)2 = (√3)2 + (√2)2 + 2√3 x √2 [ As, (a + b)2 = a2 – 2ab + b2 ]

= 3 + 2 + 2√6

= 5 + 2√6

Since, the sum of a rational and irrational number is always an irrational number.

Therefore, (√3 + √2)2 is an irrational number.

(ix) √5 – 2

Given: √5 – 2

Here, √5 is an irrational number but 2 is a rational number.

Since, the difference between an irrational number and a rational number is an irrational number.

Therefore, √5 – 2 is an irrational number.

(x) √23

Given: √23

√23 = 4.795831352331…

Since, the decimal expansion of √23 is non-terminating and non-recurring

Therefore, √23 is an irrational number.

(xi) √225

Given: √225

√225 = 15 or 15/1

Since, √225 can be represented in the form of p/q and q ≠ 0.

Therefore, √225 is a rational number

(xii) 0.3796

Given: 0.3796

Since, the decimal expansion is terminating.

Therefore, 0.3796 is a rational number.

(xiii) 7.478478……

Given: 7.478478……

Since, the decimal expansion is a non-terminating recurring decimal.

Therefore, 7.478478…… is a rational number.

(xiv) 1.101001000100001……

Given: 1.101001000100001……

Since, the decimal expansion is non-terminating and non-recurring.

Therefore, 1.101001000100001…… is an irrational number

### Question 4: Identify the following as rational or irrational numbers. Give the decimal representation of rational numbers:

(i) √4

(ii) 3√18

(iii) √1.44

(iv) √9/27

(v) – √64

(vi) √100

Solution:

(i) √4

Given: √4

Since, √4 = 2 = 2/1, it can be written in the form of a/b.

Therefore, √4 is a rational number.

The decimal representation of √4 is 2.0

(ii) 3√18

Given: 3√18

3√18 = 9√2

Since, the product of a rational and an irrational number is always an irrational number.

Therefore, 3√18 is an irrational number.

(iii) √1.44

Given: √1.44

Since, √1.44 = 1.2, it is a terminating decimal.

Therefore, √1.44 is a rational number.

The decimal representation of √1.44 is 1.2

(iv) √9/27

Given: √9/27

Since, √9/27 = 1/√3, as the quotient of a rational and an irrational number is an irrational number.

Therefore, √9/27 is an irrational number.

(v) – √64

Given: – √64

Since, – √64 = – 8 or – 8/1, as it can be written in the form of a/b.

Therefore, – √64 is a rational number.

The decimal representation of – √64 is –8.0

(vi) √100

Given: √100

Since, √100 = 10 = 10/1, as it can be written in the form of a/b.

Therefore, √100 is a rational number.

The decimal representation of √100 is 10.0

### Question 5: In the following equation, find which variables x, y, z etc. represent rational or irrational numbers:

(i) x2 = 5

(ii) y2 = 9

(iii) z2 = 0.04

(iv) u2 = 17/4

(v) v2 = 3

(vi) w2 = 27

(vii) t2 = 0.4

Solution:

(i) x2 = 5

Given: x2 = 5

When we take square root on both sides, we get,

x = √5

Since, √5 is not a perfect square root,

Therefore, x is an irrational number.

(ii) y2 = 9

Given: y2 = 9

When we take square root on both sides, we get,

y = 3

Since, 3 = 3/1, as it can be expressed in the form of a/b

Therefore, y is a rational number.

(iii) z2 = 0.04

Given: z2 = 0.04

When we take square root on both sides, we get,

z = 0.2

Since, 0.2 = 2/10, as it can be expressed in the form of a/b and is a terminating decimal.

Therefore, z is a rational number.

(iv) u2 = 17/4

Given: u2 = 17/4

When we take square root on both sides, we get,

u = √17/2

Since, the quotient of an irrational and a rational number is irrational,

Therefore, u is an irrational number.

(v) v2 = 3

Given: v2 = 3

When we take square root on both sides, we get,

v = √3

Since, √3 is not a perfect square root,

Therefore, v is an irrational number.

(vi) w2 = 27

Given: w2 = 27

When we take square root on both sides, we get,

w = 3√3

Since, the product of a rational and irrational is always an irrational number.

Therefore, w is an irrational number.

(vii) t2 = 0.4

Given: t2 = 0.4

When we take square root on both sides, we get,

t = √(4/10)

t = 2/√10

Since, the quotient of a rational and an irrational number is always an irrational number.

Therefore, t is an irrational number.

### Question 6: Give an example of each, of two irrational numbers whose:

(i) Difference in a rational number

(ii) Difference in an irrational number

(iii) Sum in a rational number

(iv) Sum is an irrational number

(v) Product in a rational number

(vi) Product in an irrational number

(vii) Quotient in a rational number

(viii) Quotient in an irrational number

Solution:

(i) Difference in a rational number

√5 is an irrational number

Since, √5 – √5 = 0

Here, 0 is a rational number.

(ii) Difference in an irrational number

Let the two irrational number be 5√3 and √3

Since, (5√3) – (√3) = 4√3

Here, 4√3 is an irrational number.

(iii) Sum in a rational number

Let the two irrational numbers be √5 and -√5

Since, (√5) + (-√5) = 0

Here, 0 is a rational number.

(iv) Sum is an irrational number

Let the two irrational numbers be 4√5 and √5

Since, 4√5 + √5 = 5√5

Here, 5√5 is an irrational number.

(v) Product in a rational number

Let the two irrational numbers be 2√2 and √2

Since, 2√2 × √2 = 2 × 2 = 4

Here, 4 is a rational number.

(vi) Product in an irrational number

Let the two irrational numbers be √2 and √3

Since, √2 × √3 = √6

Here, √6 is an irrational number.

(vii) Quotient in a rational number

Let the two irrational numbers be 2√2 and √2

Since, 2√2 / √2 = 2

Here, 2 is a rational number.

(viii) Quotient in an irrational number

Let the two irrational numbers be 2√3 and 2√2

Since, 2√3 / 2√2 = √3/√2

Here, √3/√2 is an irrational number.

### Question 7: Give two rational numbers lying between 0.232332333233332 and 0.212112111211112.

Solution:

Let a = 0.212112111211112

Let b = 0.232332333233332

Here a<b as on the second decimal place a has digit 1 and b has digit 3.

If the second decimal place is considered as 2 then it lies between a and b.

Therefore, Let x = 0.22

and y = 0.22112211…

Thus, a < x < y < b

Hence, x and y are the rational numbers required.

### Question 8: Give two rational numbers lying between 0.515115111511115 and 0.5353353335

Solution:

Let a = 0.515115111511115

Let b = 0.5353353335

Here a<b as on the second decimal place a has digit 1 and b has digit 3.

If the second decimal place is considered as 2 then it lies between a and b.

Therefore, Let x = 0.52

and y = 0.520520…

Thus, a < x < y < b

Hence, x and y are the rational numbers required.

### Question 9: Find one irrational number between 0.2101 and 0.2222…

Solution:

Let a = 0.2101

and b = 0.2222…

Here a<b as on the second decimal place a has digit 1 and b has digit 2.

If the third decimal place is considered as 1 then it lies between a and b.

Therefore, Let x = 0.2110110011…

Thus, a < x < b

Hence, x is the irrational number required.

### Question 10: Find a rational number and also an irrational number lying between the numbers 0.3030030003… and 0.3010010001…

Solution:

Let a = 0.3010010001…

and b = 0.3030030003…

Here a<b as on the third decimal place a has digit 1 and b has digit 3.

If the third decimal place is considered as 2 then it lies between a and b.

Therefore, Let x = 0.302

and y = 0.302002000200002…

Thus, a < x < y < b

Hence, x and y are the rational and irrational numbers required respectively.

### Question 11: Find two irrational numbers between 0.5 and 0.55.

Solution:

Let a = 0.5

and b = 0.55

Here a<b as on the second decimal place a has digit 0 and b has digit 5.

If the second decimal place is considered between1 to 4 then it lies between a and b.

Therefore, Let x = 0.510510051000…

and y = 0.53053530…

Thus, a < x < y < b

Hence, x and y are the irrational numbers required.

### Question 12: Find two irrational numbers lying between 0.1 and 0.12.

Solution:

Let a = 0.1

and b = 0.12

Here a<b as on the second decimal place a has digit 0 and b has digit 2.

If the second decimal place is considered 1 then it lies between a and b.

Therefore, Let x = 0.11011011000…

and y = 0.11100010100…

Thus, a < x < y < b

Hence, x and y are the irrational numbers required.

### Question 13: Prove that √3 + √5 is an irrational number.

Solution:

Let √3 + √5 be a rational number equal to x.

Therefore, x = √3 + √5

x2 = (√3 + √5)2

x2 = (√3)2 + (√5)2 + 2 √3 √5

= 3 + 5 + 2√15

= 8 + 2√15

x2 – 8 = 2√15

(x2 – 8)/2 = √15

Here, (x2 – 8)/2 is a rational but √15 is an irrational number.

Therefore, √3 + √5 is an irrational number.

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