# R.D. Sharma Solutions Class 9th: Ch 1 Number System Exercise 1.4

Chapter 1 Number System R.D. Sharma Solutions for Class 9th Math Exercise 1.4

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**Exercise 1.4**

1. Define an irrational number .

**Solution**

An irrational number is a real number that cannot be reduced to any ratio between an integer p and a natural number q. If the decimal representation of an irrational number is non-terminating and non-repeating, then it is called irrational number. For example, √3 = 1.732....

2. Explain, how irrational numbers differ from rational numbers ?

**Solution**

Every rational number must have either terminating or non-terminating but irrational number must have non- terminating and non-repeating decimal representation. A rational number is a number that can be written as simple fraction (ratio) and denominator is not equal to zero while an irrational is a number that cannot be written as a ratio.

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

**Solution**

(i) Let x = √7

Therefore,

x = 2.645751311064...

It is non-terminating and non-repeating

Hence, √7 is an irrational number

(ii) Let x = √4

Therefore,

x = 2

It is terminating.

Hence √4 is a rational number.

(iii) Let x = 2 +√3 be the rational

Squaring on both sides,

⇒ x

^{2}= (2+√3)^{2}
⇒ x

^{2}= 4+3+4√3
⇒ x

^{2}= 7+4√3
⇒ x

^{2}-7 = 4√3
⇒ (x

^{2}-7)/4 = √3
Since, x is rational

⇒ x

^{2}is rational.
⇒ x

^{2}-7 is rational
⇒ (x

^{2}-7)/4 is rational
⇒ √3 is rational

But, √3 is irrational

So, we arrive at a contradiction.

Hence 2+√3 is an irrational number.

(iv) Let x = √3+√2 be the rational number

Squaring on both sides, we get

⇒ x

^{2}= (√3+√2)^{2}
⇒ x

^{2}= 3+2+2√6
⇒ x

^{2}= 5+2√6
⇒ x

^{2}-5 = 2√6
⇒ (x

^{2}-5)/2 = √6
Since, x is a rational number

⇒ x

^{2}is rational number
⇒ x

^{2}-5 is rational number
⇒ (x

^{2}-5)/2 is rational number
⇒ √6 is rational number

But √6 is an irrational number

So, we arrive at contradiction

Hence, √3+√2 is an irrational number

(v) Let x = √3+√5 be the rational number

Squaring on both sides, we get

⇒ x

^{2}= (√3+√5)^{2}
⇒ x

^{2}= 3+5+2√15
⇒ x

^{2}= 8+2√15
⇒ x

^{2}-8 = 2√15
⇒ (x

^{2}-8)/2 = √15
Now, x is rational number

⇒ x

^{2}is rational number
⇒ x

^{2}-8 is rational number
(x

^{2}-8)/2 is rational number
⇒ √15 is rational number.

But √15 is an irrational number

So, we arrive at a contradiction

Hence, √3+√5 is an irrational number

(vi) Let x = (√2-2)

^{2}be a rational number.
x = (√2-2)

^{2}
⇒ x = 2+4-4√2

⇒ x = 6-4√2

⇒ x-6/-4 = √2

Since, x is rational number,

⇒ x – 6 is a rational nummber

⇒ x-6/-4 is a rational number

⇒ √2 is a rational number

But we know that √2 is an irrational number, which is a contradiction

So, (√2-2)

^{2}is an irrational number
(vii) Let x = (2-√2) (2+√2)

⇒ x = 2

^{2 - }√2^{2 }{as (a+b) (a-b) =^{ }2^{2 - }2^{2}}
⇒ x = 4-2

⇒ x = 2

So, (2-√2) (2+√2) is a rational number

(viii) Let x = (√2+√3)

^{2}be rational number
Using the formula, (a+b)

^{2}= 2^{2}+ 2^{2}+2ab
⇒ x = √2

^{2}+ √3^{2}+2.√2.√3
⇒ x = 2+3+2√6

⇒ x = 5+2√6

⇒ x-5/2 = √6

⇒ x-5/2 is a rational number

⇒ √6 is a rational number

But we know that √6 is an irrational number

So, we arrive at a contradiction

So, (√2+√3)

^{2}is an irrational number.
(x) Let

x = √23

⇒ x = 4.79583...

It is non-terminating or non-repeating

Hence, √23 is an irrational number

(xi) Let x = √225

⇒ x = 15

Hence √225 is a rational number

(xii) Given 0.3796

It is terminating

Hence, it is a rational number

(xiii) Given number x = 7.478478

⇒ x = 71428

It is repeating

Hence, it is a rational number

(xiv) Given number is x = 1.1010010001...

It is non-terminating or non-repeating

Hence, it is an irrational number

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

**Solution**

(i) Given number is x = √4

x = 2, which is a rational number

(iii) Given number is √1.44

Now, we have to check whether it is rational or irrational.

⇒ √1.44 = √(144/100)

⇒ √1.44 = √144/√100

⇒ √1.44 = 12/10

⇒ √1.44 = 6/5

⇒ √1.44 = 1.2

So, it is a rational

(iv) Given that √(9/27) = √9/√27

Now, we have to check whether it is rational or irrational

√(9/27) = √9/√27

⇒ √(9/27) = 3/3√3

⇒ √(9/27) = 1/√3

So, it is an irrational number

(v) Given that √-64

Now, we have to check whether it is rational or irrational

Since, -√64 = -8

So, it is a rational number

(vi) Given that √100

Now, we have to check whether it is rational or irrational

Since, √100 = 10

So, it is rational number

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

(i) x

^{2}= 5
(ii) y

^{2}= 9
(iii) z

^{2}= 0.004
(iv) u

^{2}= 17/4
(v) v

^{2}= 3
(vi) w

^{2}= 27
(vii) t

^{2}= 0.4**Solution**

(i) Given that x

^{2}= 5
Now, we have to find the value of x.

Since x

^{2}= 5
⇒ x = √5

So, it x is an irrational number

(ii) Given that y

^{2}= 9
Now, we have to find the value of y.

y

^{2}= 9
⇒ y = √9

⇒ y = 3

So, y is a rational number

(iii) Given that z

^{2}= 0.04
Now, we have to find the value of z.

⇒ z

^{2}= 4/100
⇒ z = √(4/100)

⇒ z = 2/10

⇒ z = 1/5

So, it is rational number

(iv) Given that u

^{2}= 17/4
Now, we have to find the value of u.

u

^{2}= 17/4
⇒ u = √(17/4)

⇒ u = √(17/2)

So, it is an irrational number

(v) Given that, v

^{2}= 3
Now, we have to find the value of v

v

^{2}= 3
⇒ v = √3

So, it is an irrational number

(vi) Given that, w

^{2}= 27
Now, we have to find the value of w.

⇒ w = √27

w = 3√3

So, it is an irrational number

(vii) Given that t

^{2}= 0.4
Now, we have to find the value of t

⇒ t = √0.4

⇒ t = √(4/10)

⇒ t = 2/√10

So, it is an irrational number

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

(i) difference is a rational number.

(ii) difference is an irrational number.

(iii) sum is a rational number.

(iv) sum is an irrational number.

(v) product is an rational number.

(vi) product is an irrational number.

(vii) quotient is a rational number.

(viii) quotient is an irrational number.

**Solution**

(i) Let √2, 1+√2

So, 1+√2 - √2 = 1

Therefore, √2 and 1+√2 are two irrational numbers and their difference is a rational number.

(ii) Let 4√3, 3√3 are two irrational numbers and their difference is an irrational number.

Because 4√3- 3√3 = √3 is an irrational number.

(iii) Let √5, -√5 are two irrational numbers and their sum is a rational number.

That is √5+(-√5) = 0.

(iv) Let 2√5, 3√5 are two irrational numbers and their sum is an irrational number.

That is 2√5 + 3√5 = 5√5

(v) Let √8, √2 are two irrational numbers and their product is a rational number.

That is √8×√2 = √16 = 4

(vi) Let √2, √3 are two irrational numbers and their product is an irrational number

That is √2×√3 = √6

(vii) Let √8, √2 are two irrational numbers and their quotient is a rational number

That is √8/√2 = 2√2/√2 = 2

(viii) Let √2, √3 are two irrational numbers and their quotient is an irrational number

That is √2፥√3 = √2/√3

7. Give two rational numbers lying between 0.232332333233332... and 0.212112111211112.

**Solution**

Let,

a = 0.232332333233332…..

b = 0.212112111211112…..

Here, the decimal representation of a and b are non-terminating and non-repeating. So we observe that in first decimal place of a and b have the same digit 2 but digit in the second place of their decimal representation are distinct. And the number a has 3 and b has 1. So a > b.

Hence, two rational numbers are 0.222 lying between 0.232332333233332….. and 0.212112111211112…..

8. Give two rational numbers lying between 0.515115111511115...0.5353353335...

**Solution**

Let a = 0.515115111511115… and b = 0.535335333533335….

Here, the decimal representation of a and b are non-terminating and non-repeating. So we observe that in first decimal place a and b have the same digit 5 but digit in the second place of their decimal representation are distinct. And the number a has 1 and b has 3. So a < b.

Hence, two rational numbers are 0.5152,0.532 lying between 0.515115111511115….and 0.535335333533335….

9.

**Solution**

Let,

a = 0.2101

b = 0.2222….

Here,

*a*and*b*are rational numbers .Since*a*has terminating and*b*has repeating decimal. We observe that in second decimal place*a*has 1 and*b*has 2. So*a*<*b*.
Hence one irrational number is 0.220100100010000…. lying between 0.2101 and 0.2222….

10. Find a rational number and also an irrational number lying between the numbers 0.3030030003 ... and 0.3010010001 ...

**Solution**

Let,

a = 0.3030030003…

b = 0.3010010001….

Here, decimal representation of a and b are non-terminating and non-repeating. So a and b are irrational numbers. We observe that in first two decimal place of a and b have the same digit but digit in the third place of their decimal representation is distinct. Therefore, a > b.

Hence one rational number is 0.3011 lying between 0.3030030003….. and 0.3010010001… and irrational number is 0.3020200200020000… lying between 0.3030030003….and 0.3010010001….

11. Find two irrational numbers between 0.5 and 0.55.

**Solution**

Let ,

a = 0.5

b = 0.55

Here,

*a*and*b*are rational number. So we observe that in first decimal place*a*and*b*have same digitSo a < b. Hence two irrational numbers are 0.510100100010000… and 0.5202002000200002… lying between 0.5 and 0.55.
12. Find two irrational numbers lying between 0.1 and 0.12.

**Solution**

Let,

a = 0.1

b = 0.12

Here,

*a*and*b*are rational number. So we observe that in first decimal place*a*and*b*have same digit. So a < b.
Hence, two irrational numbers are 0.1010010001… and 0.11010010001…lying between 0.1 and 0.12.

**Solution**

Given that √3+√5 is an irrational number

Now, we have to prove √3+√5 is an irrational number

Let x = √3+√5 is a rational.

Squaring on both sides,

x

^{2}= (√3 + √5)^{2}
⇒ x

^{2}= (√3)^{2}+ (√5)^{2}+ 2√3×√5
⇒ x

^{2}= 3+5+2√15
⇒ (x

^{2}= 8 + 2√15
⇒ (x

^{2}-8)/2 = √15
Now, x is rational

⇒ x

^{2}is rational
⇒ (x

^{2}-8)/2 is rational
⇒ √15 is rational

But, √15 is an irrational

Thus, we arrive at contradiction that √3+√5 is a rational which is wrong.

Hence, √3+√5 is an irrational

**Solution**

Let x = 5/7 = 0.714285 and y = 9/11 = 0.81

Here, we observe that in the first decimal x has digit 7 and y has 8. So x < y. In the second decimal place x has digit 1. So, if we considering irrational numbers

a = 0.72072007200072..

b = 0.73073007300073..

c = 0.74074007400074....

We find that,

x<a<b<c<y

Hence, a,b,c are required irrational numbers.