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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:
We know that
a^p x a^q = a^p+q
∴ Therefore

Question 2.
Solution:
We know that
a^p ÷ a^q = a^p-q
Therefore

Question 3.
Solution:
We know that
a^p x b^p = (ab)^p
Therefore

Question 4.
Solution:
We know that
(a^p)^q =a^pq

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQs
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself

Question 1.
Solution:
(b)

Its decimal will be nonterminating repeating decimal.

Question 2.
Solution:
(b) \(\frac { p }{ q }\) is terminating decimal if q = 2^m x 5ⁿ
Now, 91 = 7 x 13, 45 = 3² x 5
80 = 2⁴ x 5, 42 = 2 x 3 x 7
80 is of the form 2^m x 5ⁿ
\(\frac { 19 }{ 80 }\) is terminating decimal expansion,

Question 3.
Solution:
(b) Divisor = 9 and remainder = 7
Let b be the divisor, then
n = 9b + 7
Multiplying both sides by 3 and subtracting 1.
3n – 1 = 3(9b + 7) – 1
3n – 1 = 27b + 21 – 1
3n – 1 = 9(3b) + 9 x 2 + 2
3n – 1 = 9(3b + 2) + 2
Remainder = 2

Question 4.
Solution:
(b) \(0.\bar { 68 }\) + \(0.\bar { 73 }\)
0.686868 ……… + 0.737373……
= 1.424241 = \(1.\bar { 42 }\)

Short-Answer Questions (2 marks)
Question 5.
Solution:
4ⁿ, n ∈ N
4¹ = 4
4² = 4 x 4 = 16
4³ = 4 x 4 x 4 = 64
4⁴ = 4 x 4 x 4 x 4 = 256
4⁵ = 4 x 4 x 4 x 4 x 4 = 1024
We see that value of 4ⁿ, ends with 4 or 6 only.
Hence, the value of 4ⁿ, n ∈ N, never ends with 0.

Question 6.
Solution:
HCF of two numbers = 27 and LCM =162
One number = 81

Question 7.
Solution:
\(\frac { 17 }{ 30 }\) = \(\frac { 17 }{ 2 x 3 x 5 }\)
Here, q is in the form of 2^m x 5ⁿ
It is not terminating decimal.

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:
Let (2 + √3) is rational and 2 is rational.
Difference of them is also rational.
=> (2 + √3) – 2 = 2 + √3 – 2
= √3 is rational
But it contradicts the fact.
(2 + √3) is irrational.

Short-Answer Questions (3 marks)
Question 11.
Solution:
HCF of 12, 15, 18, 27
12 = 2 x 2 x 3 = 2² x 3
15 = 3 x 5
18 = 2 x 3 x 3 = 2 x 3²
27 = 3 x 3 x 3 = 3³
Now, HCF = 3
and LCM = 2² x 3³ x 5 =2 x 2 x 3 x 3 x 3 x 5
= 4 x 27 x 5 = 540

Question 12.
Solution:
Let 2 + √3 and 2 – √3 are two irrational number.
Sum = 2 + √3 + 2 – √3 = 4 which is a rational.

Question 13.
Solution:
4620

Question 14.
Solution:
1008

Question 15.
Solution:

Question 16.
Solution:
Give numbers are 546 and 764 and remainders are 6 and 8 respectively.
Remaining number 546 – 6 = 540
and 764 – 8 = 756
Now, required largest number = HCF of 540 and 756 = 108

Long-Answer Questions (4 marks)
Question 17.
Solution:
Let √3 is a rational number.
Let √3 = \(\frac { p }{ q }\) where p and q are integers and have no common factor, other than 1 and q ≠ 0
Squaring both sides.
3 = \(\frac { { p }^{ 2 } }{ { q }^{ 2 } }\) => 3q² – p²
=> 3 divides p²
=> 3 divides p
Let p = 3c for some integer c
3q² = 9c² => q² – 3c²
=> 3 divides q² (3 divides 3c²)
=> 3 divides q
3 is common factors of p and q
But it contradicts the fact that p and q have
no common factors and also contradicts that √3 is a rational number.
Hence, √3 is irrational number.

Question 18.
Solution:
Let n be an arbitrary odd positive integer on dividing n by 4, let m be the quotient and r be the remainder.
By Euclid’s division lemma,
n = 4q + r where 0 ≤ r < 4
n = 4q or (4q + 1) or (4q + 2) or (4q + 3)
Clearly, 4q and (4q + 2) are even number
since n is odd.
n ≠ 4q and n ≠ (4q + 2)
n = (4 q + 1) or (4q + 3) for same integer n
Hence, any positive odd integer of the form 4q + 1 or 4q + 3 for some integer q.

Question 19.
Solution:
On dividing n by 3, let q be the quotient and r be the remainder, then
n = 3q + r where 0 ≤ r < 3 => n = 3q + r where r = 0, 1 or 2
n = 3q or n = 3q + 1 or n = 3q + 2
(i) Case (I)
If n = 3q then n is divisible by 3
(ii) Case (II)
If n = (3q + 1) then n + 2 = 3q + 3 = 3q (q + 1) which is divisible by 3
In this case, n + 2 is divisible by 3
(iii) Case (III)
If n = (3q + 2) then n + 1 (n + 1) = 3q + 3 = 3(q + 1) which also divisible by 3
In this case, (n + 1) is divisible by 3
Hence, one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

Question 20.
Solution:
Let (4 + 3√2) is rational number and 4 is also a rational number.
Difference of two rational numbers is also a rational number.
4 + 3√2 – 4 = 3√2 is a rational number
Product of two rational numbers is rational
3 is rational and √2 is rational
But it contradicts the fact
√2 is irrational
Hence, (4 + 3√2 ) is irrational.

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself  are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers MCQs
• RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Test Yourself

Very-Short Answer Questions
Question 1.
Solution:
For any two given positive integers a and b there exist unique whole numbers q and r such that
a = bq + r, where 0 ≤ r < b.
Here, we call ‘a’ as dividend, b as divisor, q is quotient and r as remainder.
Dividend = (Divisor x Quotient) + Remainder

Question 2.
Solution:
Every composite number can be uniquely expressed as a product of two primes, except for the order in which these prime factors occurs.

Question 3.
Solution:
360 = 2 x 2 x 2 x 3 x 3 x 5 = 2³ x 3² x 5

Question 4.
Solution:
We know that HCF of two primes is
HCF (a, b) = 1

Question 5.
Solution:
a and b are two prime numbers then their
LCM = Product of these two numbers
LCM(a, b) = a x b = ab.

Question 6.
Solution:
We know that product of two numbers is equal to their HCF x LCM
LCM = \(\frac { Product of two numbers }{ HCF }\)
= \(\frac { 1050 }{ 25 }\) = 42
LCM of two numbers = 42

Question 7.
Solution:
A composite number is a number which is not a prime. In other words, a composite number has more than two factors.

Question 8.
Solution:
a and b are two primes, then their
HCF will be 1
HCF of a and b = 1

Question 9.
Solution:
\(\frac { a }{ b }\) is a rational number and it has terminating decimal
b will in the form 2^m x 5ⁿ where m and n are some non-negative integers.

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
2ⁿ x 5ⁿ = (2 x 5)ⁿ = (10)ⁿ
Which always ends in a zero
There is no value of n for which (2ⁿ x 5ⁿ) ends in 5

Question 13.
Solution:
We know that HCF is always a factor is its LCM
But 25 is not a factor of 520
It is not possible to have two numbers having HCF = 25 and LCM = 520

Question 14.
Solution:
Let two irrational number be (5 + √3) and (5 – √3).
Now their sum = (5 + √3) + (5 – √3) = 5 + √3 + 5 – √3 = 10
Which is a rational number.

Question 15.
Solution:
Let the two irrational number be (3 + √2) and (3 – √2)
Now, their product = (3 + √2) (3 – √2)
= (3)² – (√2)² {(a + b) (a – b) = a² – b²}
= 9 – 2 = 7
Which is a rational number.

Question 16.
Solution:
a and b are relative primes
their HCF = 1

Question 17.
Solution:
LCM of two numbers = 1200
and HCF = 500
But we know that HCF of two numbers divides their LCM.
But 500 does not divide 1200 exactly
Hence, 500 is not their HCF whose LCM is 1200.

Short-Answer Questions
Question 18.
Solution:
Let x = 0.4 = 0.444
Then 10x = 4.444….
Subtracting, we get
9x = 4 => x = \(\frac { 4 }{ 9 }\)
\(\bar { 0.4 }\) = \(\frac { 1 }{ 2 }\) which is in the simplest form.

Question 19.
Solution:
\(\bar { 0.23 }\)
Let x = \(\bar { 0.23 }\) = 0.232323…….
and 100x = 23.232323……
Subtracting, we get
99x = 23 => x = \(\frac { 23 }{ 99 }\)
\(\bar { 0.23 }\) = \(\frac { 23 }{ 99 }\) which is in the simplest form.

Question 20.
Solution:
0.15015001500015
It is non-terminating non-repeating decimal.
It is an irrational number.

Question 21.
Solution:
\(\frac { \surd 2 }{ 3 }\) = \(\frac { 1 }{ 3 }\) √2
Let \(\frac { 1 }{ 3 }\) √2 is a rational number
Product of two rational numbers is a rational
\(\frac { 1 }{ 3 }\) is rational and √2 is rational contradicts
But it contradicts the fact
\(\frac { \surd 2 }{ 3 }\) or \(\frac { 1 }{ 3 }\) √2 is irrational.

Question 22.
Solution:
√3 and 2.
√3 = 1.732 and 2.000
A rational number between 1.732 and 2.000 can be 1.8 or 1.9
Hence, 1.8 or 1.9 is a required rational.

Question 23.
Solution:
\(\bar { 3.1416 }\)
It is non-terminating repeating decimal.
It is a rational number.

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Rationalise the denominator of each of the followings :

Question 1.
Solution:
Here,RF of √7 is √7

Question 2.
Solution:
Here RF √3 is √3

Question 3.
Solution:
Here RF of \(\frac { 1 }{ \left( { 2+ }\sqrt { 3 } \right) }\) is \(\frac { 1 }{ \left( { 2- }\sqrt { 3 } \right) }\)

Question 4.
Solution:
Here RF is √5 + 2

Question 5.
Solution:
Here RF is 5 – 3√2

Question 6.
Solution:
Here RF is √6 + √5

Question 7.
Solution:
Here RF = √7 – √3

Question 8.
Solution:
Here RF = √3 – 1

Question 9.
Solution:
Here RF = (3-2√2)

Find the values of a and b in each of the following :

Question 10.
Solution:
\(\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \), RF = √3+1
(Multiplying and dividing by √3+1)

Question 11.
Solution:
\(\frac { 3+\sqrt { 2 } }{ 3-\sqrt { 2 } } \), RF is 3+√2
(Multiplying and dividing by 3+√2)

Question 12.
Solution:
In \(\frac { 5-\sqrt { 6 } }{ 5+\sqrt { 6 } } \), RF is (5-√6)
(Multiplying and dividing by 5-√6)

Question 13.
Solution:
In \(\frac { 5+2\sqrt { 3 } }{ 7+4\sqrt { 3 } } \), RF is 7-4√3
(Multiplying and dividing by 7-4√3)

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
x = (4-√15)

Question 17.
Solution:
x = 2+√3

Question 18.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
(i) Draw a line segment AB = 3.2 units (cm) and extend it to C such that BC = 1 unit.

(ii) Find the mid-point O of AC.
(iii) With centre O and OA as radius draw a semicircle on AC
(iv) Draw BD ⊥ AC meeting the semicircle at D.
(v) Join BD which is √3.2 units.
(vi) With centre B and radius BD, draw an arc meeting AC when produced at E.
Then BE = BD = √3.2 units. Ans.

Question 6.
Solution:
(i) Draw a line segment AB = 7.28 units and produce is to C such that BC = 1 unit (cm)

(ii) Find the mid-point O of AC.
(iii) With centre O and radius OA, draw a semicircle on AC.
(iv) Draw a perpendicular BD at AC meeting the semicircle at D
Then BD = √7.28 units.
(v) With centre B and radius BD, draw an arc which meet AC produced at E.
Then BE = BD = √7.28 units.

Question 7.
Solution:
(A) For Addition
(i) Closure property: The sum of two real numbers is always a real number.
(ii) Associative Law : (a + b) + c = a + (b + c), for all values of a, b and c.
(iii) Commutative Law : a + b = b + a for all real values of a and b.
(iv) Existance of Additive Identity : 0 is the real number such that: 0 + a = a + 0 = afor every real value of a.
(v) Existance of addtive inverse : For each real value of a, there exists a real value (-a) such that a + (-a) = (-a) + a = 0, Then (a) and (-a) are called the additive inverse of each other.
(v) Existence of Multiplicative Inverse. For each non zero real number a, there exists a real number \(\frac { 1 }{ a }\) such that a . \(\frac { 1 }{ a }\) = \(\frac { 1 }{ a }\) . a = 1
a and \(\frac { 1 }{ a }\) are called multiplicative inverse or reciprocal of each other.
(B) Multiplication
(i) Closure property: The product of two real numbers is always a real number.
(ii) Associative law : ab(c) = a(bc) for all real values of a, b and c
(iii) Commutative law : ab=ba for all real numbers a and b
(iv) Existance of Multiplicative Identity: clearly is a real number such that 1.a = a.1 = a for every value of a.

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:
Irrational numbers : Numbers which are not rational numbers, are called irrational numbers. Rational numbers can be expressed in terminating decimals or repeating decimals while irrational number can’t.
\(\frac { 1 }{ 2 } \) , \(\frac { 2 }{ 3 } \) , \(\frac { 7 }{ 5 } \) etc.are rational numbers and π, √2, √3, √5, √6….etc are irrational numbers

Question 2.
Solution:
(i) √4 = ±2, it is a rational number
(ii) √196 = ±14 it is a rational number
(iii) √21 It is irrational number.
(iv) √43 It is irrational number.
(v) 3 + √3 It is irrational number because sum of a rational and an irrational number is irrational
(vi) √7 – 2 It is irrational number because difference of a rational and irrational number is irrational
(vii) \(\frac { 2 }{ 3 } \)√6 . It is irrational number because product of a rational and an irrational number is an irrational number.
(viii) 0.\(\overline { 6 } \) = 0.6666…. It is rational number because it is a repeating decimal.
(ix) 1.232332333…. It is irrational number because it not repeating decimal
(x) 3.040040004…. It is irrational number because it is not repeating decimal.
(xi) 3.2576 It is rational number because it is a terminating decimal.
(xii) 2.3565656…. = 2.3 \(\overline { 56 } \) It is rational number because it is a repeating decimal.
(xiii) π It is an irrational number
(xiv) \(\frac { 22 }{ 7 } \). It is a rational number which is in form of \(\frac { p }{ q } \) Ans.

Question 3.
Solution:
(i) Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.
Taken OA = 1 unit and draw AB ⊥ OA such that AB = 1 unit. Join OB, Then,

Question 4.
Solution:
Firstly we represent √5 on the real line X’OX. Then we will find √6 and √7 on that real line.
Now, draw a horizontal line X’OX, taken as x-axis

Question 5.
Solution:
(i) 4 + √5 : It is irrational number because in it, 4 is a rational number and √5 is irrational and sum of a rational and an irrational is also an irrational.
(ii) (-3 + √6) It is irrational number because in it, -3 is a rational and √6 is irrational and sum or difference of a rational and irrational is an irrational.
(iii) 5√7 : It is irrational because 5 is rational and √7 is irrational and product of a rational and an irrational is an irrational.
(iv) -3√8 : It is irrational because -3 is a rational and √8 is an irrational and product of a rational and an irrational is also an irrational.
(v) \(\frac { 2 }{ \sqrt { 5 } } \) It is irrational because 2 is a rational and √5 is an irrational and quotient of a rational and an irrational is also an irrational.
(vi) \(\frac { 4 }{ \sqrt { 3 } } \) It is irrational because 4 is a rational and √3 is an irrational number and quotient of a rational and irrational is also an irrational.

Question 6.
Solution:
(i) True.
(ii) False, as the sum of two irrational number is irrational is not always true.
(iii) True.
(iv) False, as the product of two irrational numbers is irrational is not always true.
(v) True.
(vi) True.
(vii) False as a real number can be either rational or irrational.

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Question 1.
Solution:
(i) Rational numbers: Numbers in the form of \(\frac { p }{ q }\) where p and q are integers and q ≠ 0, are called rational numbers.
(ii) Irrational numbers : The numbers which are not rationals, are called irrational numbers. Irrational numbers can be expressed in decimal form as non terminating non-repeating decimal.
(iii) Real numbers : The numbers which are rational or irrational, are called real numbers.

Question 2.
Solution:
(i) \(\frac { 22 }{ 7 }\)
It is a rational number as it is in the form of \(\frac { p }{ q }\)
(ii) 3.1416
It is a rational number as it is a terminating decimal.
(iii) π
It is an irrational number as it is nonterminating non-repeating decimal.
(iv) \(3.\bar { 142857 }\)
It is a rational number as it is nonterminating repeating decimal.
(v) 5.636363… = 5.63
It is a rational number as it is nonterminating repeating decimal.
(vi) 2.040040004…
It is an irrational number as it is nonterminating non-repeating decimal.
(vii) 1.535335333…
It is an irrational number as it is non terminating non-repeating decimal.
(viii) 3.121221222…
It is an irrational number as it is nonterminating non-repeating decimal.
(ix) √21
It is an irrational number aS it is not in the form of \(\frac { p }{ q }\)
(x) \(\sqrt [ 3 ]{ 3 }\)
It is an irrational number as it is not in the form of \(\frac { p }{ q }\)

Question 3.
Solution:
(i) √6 is irrational.
Let √6 is not an irrational number, but it is a rational number in the simplest form of \(\frac { p }{ q }\)
√6 = \(\frac { p }{ q }\) (p and q have no common factors)
Squaring both sides,
6 = \(\frac { { p }^{ 2 } }{ { q }^{ 2 } }\)
p² = 6q²
p² is divisible by 6
=> p is divisible by 6
Let p = 6a for some integer a
6q² = 36a²
=> q² = 6a²
q² is also divisible by 6
=> q is divisible by 6
6 is common factors of p and q
But this contradicts the fact that p and q have no common factor
√6 is irrational
(ii) (2 – √3) is irrational
Let (2 – √3) is a rational and 2 is also rational, then
2 – (2 – √3 ) is rational (Difference two rationals is rational)
=> 2 – 2 + √3 is rational
=> √3 is rational
But it contradicts the fact
(2 – √3) is irrational
(iii) (3 + √2 ) is irrational
Let (3 + √2 ) is rational and 3 is also rational
(3 + √2 ) – 3 is rational (Difference of two rationals is rational)
=> 3 + √2 – 3 is rational
=> √2 is rational
But it contradicts the fact (3 + √2 ) is irrational
(iv) (2 + √5 ) is irrational
Let (2 + √5 ) is rational and 2 is also rational
(2 + √5) – 2 is rational (Difference of two rationals is rational)
=> 2 + √5 – 2 is rational
=> √5 is rational
But it contradicts the fact (2 + √5) is irrational
(v) (5 + 3√2 ) is irrational
Let (5 + 3√2 ) is rational and 5 is also rational
(5 + 3√2 ) – 5 is rational (Difference of two rationals is rational)
=>5 + 3√2 – 5 is rational
=> 3√2 is rational
Product of two rationals is rational
3 is rational and √2 is rational
√2 is rational
But it contradicts the fact
(5 + 3√2 ) is irrational
(vi) 3√7 is irrational
Let 3√7 is rational
3 is rational and √7 is rational (Product of two rationals is rational)
But √7 is rational, it contradicts the fact
3√7 is irrational
(vii) \(\frac { 3 }{ \surd 5 }\) is irrational
Let \(\frac { 3 }{ \surd 5 }\) is rational
\(\frac { 3\times \surd 5 }{ \surd 5\times \surd 5 } =\frac { 3\surd 5 }{ 5 }\) is rational
\(\frac { 3 }{ 5 }\) is rational and √5 is rational
But √5 is a rational, it contradicts the fact
\(\frac { 3 }{ \surd 5 }\) is irrational
(viii)(2 – 3√5) is irrational
Let 2 – 3√5 is rational, 2 is also rational
2 – (2 – 3√5) is rational (Difference of two rationals is rational)
2 – 2 + 3√5 is rational
=> 3√5 is rational
3 is rational and √5 is rational (Product of two rationals is rational)
√5 is rational
But it contradicts the fact
(2 – 3√5) is irrational
(ix) (√3 + √5) is irrational
Let √3 + √5 is rational
Squaring,
(√3 + √5)² is rational
=> 3 x 5 + 2√3 x √5 is rational
=> 8 + 2√15 is rational
=> 8 + 2√15 – 8 is rational (Difference of two rationals is rational)
=> 2√15 is rational
2 is rational and √15 is rational (Product of two rationals is rational)
√15 is rational
But it contradicts the fact
(√3 + √5) is irrational

Question 4.
Solution:
Let \(\frac { 1 }{ \surd 3 }\) is rational
= \(\frac { 1 }{ \surd 3 } \times \frac { \surd 3 }{ \surd 3 } =\frac { \surd 3 }{ 3 } = \frac { 1 }{ 3 } \surd 3\) is rational
\(\frac { 1 }{ 3 }\) is rational and √3 is rationals (Product of two rationals is rational)
√3 is rational But it contradicts the fact
\(\frac { 1 }{ \surd 3 }\) is irrational

Question 5.
Solution:
(i) We can take two numbers 3 + √2 and 3 – √2 which are irrationals
Sum = 3 + √2 + 3 – √2 = 6 Which is rational
3 + √2 and 3 – √2 are required numbers
(ii) We take two. numbers
5 + √3 and 5 – √3 which are irrationals
Now product = (5 + √3) (5 – √3)
= (5)² – (√3 )² = 25 – 3 = 22 which is rational
5 + √3 and 5 – √3 are the required numbers

Question 6.
Solution:
(i) True.
(ii) True.
(iii) False, as sum of two irrational can be rational number also such as
(3 + √2) + (3 – √2) = 3 + √2 + 3 – √2 = 6 which is rational.
(iv) False, as product of two irrational numbers can be rational also such as
(3 + √2)(3 – √2 ) = (3)2 – (√2 )2 = 9 – 2 = 7
which is rational
(v) True.
(vi) True.

Question 7.
Solution:
Let (2√3 – 1) is a rational number and 1 is a rational number also.
Then sum = 2√3 – 1 + 1 = 2√3
In 2√3, 2 is rational and √3 is rational (Product of two rational numbers is rational)
But √3 is rational number which contradicts the fact
(2√3 – 1) is an irrational.

Question 8.
Solution:
Let 4 – 5√2 is a rational number and 4 is also a rational number
Difference of two rational number is a rational numbers
4 – (4 – 5√2 ) is rational
=> 4 – 4 + 5√2 is rational
=> 5√2 is rational
Product of two rational number is rational
5 is rational and √2 is rational
But it contradicts the fact that √2 is rational √2 is irrational
Hence, 4 – 5√2 is irrational

Question 9.
Solution:
Let (5 – 2√3) is a rational number and 5 is also a rational number
Difference of two rational number is rational
=> 5 – (5 – 2√3) is rational
=> 5 – 5 + 2√3 or 2√3 is rational
Product of two rational number is rational
2 is rational and √3 is rational
But it contradicts the fact
(5 – 2√3) is an irrational number.

Question 10.
Solution:
Let 5√2 is a rational
Product of two rationals is a rational
5 is rational and √2 is rational
But it contradicts the fact
5√2 is an irrational.

Question 11.
Solution:
\(\frac { 2 }{ \surd 7 } =\frac { 2\surd 7 }{ \surd 7\times \surd 7 } =\frac { 2\surd 7 }{ 7 } =\frac { 2 }{ 7 } \surd 7\)
Let \(\frac { 2 }{ 7 } \surd 7\) is a rational number, then
\(\frac { 2 }{ 7 }\) is rational and √7 is rational
But it contradicts the fact \(\frac { 2 }{ \surd 7 }\) is an irrational number.

Hope given RS Aggarwal Solutions Class 10 Chapter 1 Real Numbers Ex 1D are helpful to complete your math homework.

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