## 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.

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