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

**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:

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)√7Given: √7

Since, it is not a perfect square root,

Therefore, it is an irrational number.

(ii) √4Given: √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 + √3Given: 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 + √2Given: √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 + √5Given: √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) = a^{2}– b^{2}]= 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}= a^{2}– 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 – 2Given: √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)√23Given: √23

√23 = 4.795831352331…

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

Therefore, √23 is an irrational number.

(xi)√225Given: √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) 0.3796Given: 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) √4Given: √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√18Given: 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)√1.44Given: √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)√9/27Given: √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)– √64Given: – √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)√100Given: √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) x ^{2} = 5**

**(ii) y2 = 9**

**(iii) z ^{2} = 0.04**

**(iv) u ^{2 }= 17/4**

**(v) v ^{2} = 3**

**(vi) w ^{2} = 27**

**(vii) t ^{2} = 0.4**

**Solution:**

(i) x^{2}= 5Given: x

^{2}= 5When 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)y^{2}= 9Given: y

^{2}= 9When 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)z^{2}= 0.04Given: z

^{2}= 0.04When 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) u^{2}= 17/4Given: u

^{2}= 17/4When 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)v^{2}= 3Given: v

^{2}= 3When 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) w^{2}= 27Given: w

^{2}= 27When 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)t^{2}= 0.4Given: t

^{2}= 0.4When 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 numberLet 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 numberLet the two irrational numbers be √5 and -√5

Since, (√5) + (-√5) = 0

Here, 0 is a rational number.

(iv) Sum is an irrational numberLet 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 numberLet 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 numberLet the two irrational numbers be √2 and √3

Since, √2 × √3 = √6

Here, √6 is an irrational number.

(vii) Quotient in a rational numberLet 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 numberLet 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

ahas digit 1 andbhas digit 3.If the second decimal place is considered as 2 then it lies between

aandb.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

ahas digit 1 andbhas digit 3.If the second decimal place is considered as 2 then it lies between

aandb.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

ahas digit 1 andbhas digit 2.If the third decimal place is considered as 1 then it lies between

aandb.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

ahas digit 1 andbhas digit 3.If the third decimal place is considered as 2 then it lies between

aandb.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

x

^{2}= (√3 + √5)^{2}x

^{2}= (√3)^{2}+ (√5)^{2}+ 2 √3 √5= 3 + 5 + 2√15

= 8 + 2√15

x

^{2}– 8 = 2√15(x

^{2}– 8)/2 = √15Here, (x

^{2}– 8)/2 is a rational but √15 is an irrational number.Therefore, √3 + √5 is an irrational number.

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