Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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: 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :