Chapter 1 Number System RD Sharma Solutions Exercise 1.4 Class 9 Maths
Chapter Name  RD Sharma Chapter 1 Number System Exercise 1.4 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 1.4 Solutions
1. Define an irrational number.
Solution
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number. For example, 1.01001000100001......
2. Explain, how irrational numbers differ from rational numers ?
Solution
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number, for example, and 6.2876 are rational numbers.
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) (√2 + √3)^{2 }
(ix) √5  2
(x) √23
(xi) √225
(xii) 0.3796
(xiii) 7.478478
(xiv) 1.101001000100001.....
Solution
√7 is not a perfect square root, so it is an irrational number.
We have,
√4 = 2 = 2/1
∴ √4 can be expressed in the form of p/q, so it is a rational number.
The decimal representation of √4 is 2.0 .
2 is rational number, whereas √3 is an irrational number.
Because, sum of a rational number and an irrational number is an irrational number, so 2 + √3 is an irrational number.
√2 is an irrational number. Also √3 is an irrational number.
The sum of two irrational numbers is irrational.
∴ √3 + √2 is an irrational number.
√5 is an irrational number. Also √3 is an irrational number.
The sum of two irrational numbers is irrational.
∴ √3 + √5 is an irrational number.
We have,
(√2  2)^{2 } = (√2)^{2 }  2×√2 ×2 + (2)^{2 }
= 2  4√2 + 4
= 6  4√2
Now, 6 is a rational number, whereas 4√2 is an irrational number.
The difference of a rational number and an irrational number is an irrational number.
So, 6  4√2 is an irrational number.
∴ (√2  2)^{2 } is an irrational number.
We have,
(2  √2)(2 + √2) = (2)^{2 }  (√2)^{2 }
= 4  2
= 2 = 2/1
Since, 2 is a rational number.
∴ (2  √2) (2 + √2) is a rational number.
We have,
(√2 + √3)^{2 } = (√2)^{2 } + 2× √2 × √3 + (√3)^{2 }
= 2 + 2√6 + 3
= 5 + 2√6
The sum of a rational number and an irrational number is an irrational number, so 5 + 2√6 is an irrational number.
∴ (√2 + √3)^{2} is an irrational number.
The difference of a rational number and an irrational number is an irrational number.
∴ √5  2 is an irrational number.
√23 = 4.7958312331......
√225 = 15 = 15/1
Rational number as it can be represented in p/q form .
0.3796
As decimal expansion of this number is terminating, so it is a rational number.
7.478478 .... =
As decimal expansion of this number is non  terminating recurring so it is a rational number.
4. Identify the following as rational 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) We have
√4 = 2 = 2/1
√4 can be written in the form of p/q, so it is a rational number.
Its decimal representation is 2.0.
(ii) We have,
3√18 = 3√(2×3×3)
= 3 ×3√2
= 9√2
Since, the product of a rations and an irrational is an irrational number.
∴ 9√2 is an irrational
⇒ 3√18 is an irrational number.
(iii) We have,
√1.44 = √(144/100)
= 12/10 = 1.2
Every terminating decimal is a rational number, so 1.2 is a rational number.
Its decimal representation is 1.2.
(iv) We have,
√(9/27)
= 3/√27
= 3/(√3 ×3×3)
= 3/3√3
= 1/√3
Quotient of a rational and an irrational number is irrational numbers so 1/√3 is an irrational number.
⇒ √(9/27) is an irrational number.
(v) We have,
√6
= √(8 ×8)
= 8
= 8/1
√64 can be expressed in the form of p/q , so √64 is a rotational number.
Its decimal representation is 8.0.
(vi) We have,
√100 = 10
= 10/1
√100 can be expressed in the form of p/q. So √100 is a rational number.
The decimal representation of √100 is 10.0.
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 } = 0.04
(iv) u^{2 } = 174
(v) v^{2 } = 3
(vi) w^{2 } = 27
(vii) t^{2 } = 0.4
Solution
(i) We have
x^{2 } = 5
Taking square root on both sides.
⇒ √x^{2 } = √5
⇒ x = √5
√5 is not a perfect square root, so it is an irrational number.
(ii) We have
y^{2 } = 9
⇒ y = √9
= 3
= 3/1
√9 can be expressed in the form of p/q , so it a rational number.
(iii) We have
z^{2 } = 0.04
Taking square root on the both sides, we get,
√z^{2 } = √0.04
⇒ z = √0.04
= 0.2
= 2/10
= 1/5
z can be expressed in the form of p/q, so it is a rational number.
(iv) We have
u^{2 } = 17/4
Taking square root on both sides, we get,
√u^{2 } = √(17/4)
⇒ u = √(17/2)
Quotient of an irrational and a rational number is irrational, so u is an irrational number.
(v) We have
v^{2 } = 3
Taking square root on both sides, we get,
√v^{2 } = √13
⇒ v = √3
√3 is not a perfect square root, so y is an irrational number.
(vi) We have
w^{2 } = 27
Taking square root on both des, we get,
√w^{2 } = √27
⇒ w = √3 × 3×3
= 3√3
Product of a rational and an irrational is irrational number, so w is an irrational number.
(vii) We have
t^{2 } = 0.4
Taking square root on both sides, we get
√t^{2 } = √0.4
⇒ t = √4/10
= 2/√10
Since, quotient of a rational and an irrational number is irrational number, so t is an irrational number.
6. Given 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 a rational number.
(vi) product is an irrational number.
(vii) quotient is a rational number.
(viii) quotient is an irrational number.
Solution
(i) √3 is an irrational number.
Now, (√3)  (√3) = 0
0 is the rational number.
(ii) Let two irrational numbers are 5√2 and √2
Now, (5√2)  (√2) = 4√2
4√2 is the rational number.
(iii) Let two irrational numbers are √11 and √11
Now, (√11) +(√11) = 0
0 is the rational number.
(iv) Let two irrational numbers are 4√6 and √6
Now, (4√6) + (√6) = 5√6
5√6 is the rational number.
(v) Let two irrational numbers are 2√3 and √3
Now, 2√3 × √3 = 2×3 = 6
6 is the rational number.
(vi) Let two irrational numbers are √2 and √5
Now, √2 × √5 = √10
√10 is the rational number.
(vii) Let two irrational numbers are 3√6 and √6
Now, (3√6)/√6 = 3
3 is the rational number.
(viii) Let two irrational numbers are √6 and √2
Now, √6/√2 = [√(3+2)/√2
= (√3 × √2)/√2
= √3
√3 is an irrational number.
7. Give two rational numbers lying between 0.232332333233332...... and 0.212112111211112.
Solution
Let, a = 0.212112111211112
And, b = 0.232332333233332......
Clearly, a < b because in the second decimal place a has digit 1 and b has digit 3
If we consider rational numbers in which the second decimal place has the digit 2, then they will lie between a and b.
Let,
x = 0.22
y = 0.22112211.....
then,
a < x < y < b
Hence, x, and y are required rational numbers.
8. Give two rational numbers lying between 0.515115111511115 ....... and 0.5353353335.....
Solution
Let, a = 0.515115111511115 .....
And, b = 0.5353353335 .....
We observe that in the second decimal place a has digit 1 and b has digit 3, therefore, a < b. So, if we consider rational numbers
x = 0.52
y = 0.52052052 .....
We find that,
a < x < y < b
Hence x, and y are required rational numbers.
9. Find one irrational number between 0.2101 and 0.2222 ..... = 0.2
Solution
Let, a = 0.2101
And, b = 0.2222 .....
We observe that in the second decimal place a has digit 1 and b has digit 2, therefore a < b in the third decimal place a has digit 0. So, if we consider irrational numbers
x = 0.211011001100011.....
We find that
a < x < b
Hence, x is required irrational number.
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....
We observe that in the third decimal place a has digit 1 and b has digit 3, therefore a < b, in the third decimal place a has digit 1. So if we consider rational and irrational numbers
x = 0.302
y = 0.302002000200002 .....
We find that
a < x < b
And, a < y < b
Hence, x and y are required rational and irrational numbers respectively.
11. Find two irrational numbers between 0.5 and 0.55.
Solution
Let a = 0.5 = 0.50
And, b = 0.55
We observe that in the second decimal place a has digit 0 and b has digit 5 , therefore a < b.
so, if we consider irrational numbers
x = 0.51051005100051....
y = 0.530535305353530......
We find that
a < x < y < b
Hence, x and y are required irrational numbers.
12. Find two irrational numbers lying between 0.1 and 0.12.
Solution
Let, a = 0.1 = 0.10
And, b = 0.12
We observe that in the second decimal place a has digit 0 and b has digit 2, Therefore a < b. So, if we consider irrational numbers
x = 0.11011001100011....
y = 0.111011110111110.....
We find that,
a < x < y < b
Hence, x and y are required irrational numbers.
13. Prove that √3 + √5 is an irrational number.
Solution
If possible, let √3 + √5 be a rational number equal to x. Then,
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 = √15
Now, x is rational
⇒ x^{2 } is rational
⇒ (x^{2 }  8)/2 is rational
⇒ √15 is rational
But, √15 is rational
Thus, we arrive at a contradiction. So, our supposition that √3 + √5 is rational is wrong.
Hence, √3 + √5 is an irrational number.
14. Find three different irrational numbers between the rational numbers 5/7 and 9/11
Solution
5/7 = 0. 714285
9/11 = 0. 81
3 irrational numbers are
0.73073007300073000073 ......
0.75075007500075000075 .....
0.79079007900079000079 .....