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Chapter 1 of RD Sharma’s Class 9 Mathematics textbook covers the fundamentals of the Number System. Exercise 1.4 delves into specific problems and applications related to the various types of numbers including integers, rational numbers, and irrational numbers. The problems are designed to enhance students' understanding of the properties and operations involving these numbers. This exercise is crucial for building a strong foundation in mathematical reasoning and problem-solving.
The number system is a way of representing and categorizing numbers based on their properties and relationships. It includes different types of numbers: natural numbers, whole numbers, integers, rational numbers, and irrational numbers. Each type has unique characteristics and operations. Understanding the number system is fundamental for performing arithmetic operations and solving mathematical problems efficiently.
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.....
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
(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 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 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
√3 is not a perfect square, thus it is an irrational number.
√2 is also not a perfect square, thus it is an irrational number.
√3+√2 cannot be simplified to a rational number since there is no way to express it as the ratio of two integers (a rational number).
The sum of √3+√2 is not a perfect square and does not cancel out any irrational terms. Hence, it remains irrational.
(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.
√3+√5 cannot be simplified to a rational number since there is no way to express it as the ratio of two integers (a rational number).
The sum of √3+√5 is not a perfect square and does not cancel out any irrational terms. Hence, it remains irrational.
(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, ]
= 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
(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
(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.
(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.
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.
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.
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.
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.
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.
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.
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.
Read More:
Exercise 1.4 in RD Sharma's Class 9 textbook focuses on the important concept of rationalizing denominators, particularly those involving surds. This skill is crucial for simplifying complex expressions and is widely used in algebra and calculus. By mastering these techniques, students develop a deeper understanding of irrational numbers and learn to manipulate them effectively in mathematical expressions.
