RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers, contains solutions for all Exercise 1D questions. This exercise is based on concepts such as Irrational Numbers and some of its important properties. This set of questions and answers is prepared by BYJU’S subject experts to help students to do well in their exams. Download the RS Aggarwal Solutions of Class 10 now and practice all the questions.

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Exercise 1A Solutions : 10 Questions (Short Answers)

Exercise 1B Solutions : 27 Questions (Short Answers)

Exercise 1C Solutions : 3 Questions (Short Answers)

Exercise 1E Solutions : 23 Questions (Short Answers)

## Exercise 1D Page No: 33

**Question 1: Define (i) rational numbers, (ii) irrational numbers, (iii) real numbers.**

**Solution**:

Rational numbers: The numbers of the form p/q , where p and q are integers and q â‰ 0.

Irrational numbers: The numbers which when expressed in decimal form and expressible as non-terminating and non-repeating decimals.

Real numbers: Combination of rational and irrational numbers.

**Question 2: Classify the following numbers as rational or irrational:**

**Solution**:

(i) 22/7 is a rational number.

(ii) 3.1416 is a rational number.

It is a terminating decimal and non-repeating decimal.

(iii) Ï€ is an irrational number.

It is a non-terminating and non-repeating decimal.

(iv) \(3.\overline{142857}\)

A rational number. Non-terminating repeating decimal.

(v) 5.636363…

A rational number. A non-terminating repeating decimal.

(vi) 2.040040004…

An irrational number. It is a non-terminating and non-repeating decimal.

(vii) 1.535335333…

An irrational number. A non-terminating and non-repeating decimal.

(viii) 3.121221222…

An irrational number. A non-terminating and non-repeating decimal.

(ix)\(\sqrt{21}\)

An irrational number.

21 = 3 Ã— 7 is an irrational number. And 3 and 7 are prime and irrational numbers.

(x)\(\sqrt[3]{3}\)

An irrational number.

3 is a prime number. So, 3 is an irrational number.

**Question 3: Prove that each of the following numbers is irrational.**

**Solution:**

**(i) âˆš6**

Let us suppose that âˆš6 is a rational number.

There exists two co-prime numbers, say p and q

So, âˆš6 = p/q

Squaring both sides, we get

6 = p^{2}/q^{2}

or 6q^{2} = p^{2} â€¦(1)

Which shows that, pÂ² is divisible by 6

This implies, p is divisible by 6

Let p = 6a for some integer a

Equation (1) implies: 6qÂ² = 36aÂ²

â‡¨ qÂ² = 6aÂ²

qÂ² is also divisible by 6

â‡¨ 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

Our assumption is wrong. Thus, âˆš6 is irrational

**(ii) (2 â€“ âˆš3)**

Let us assume that (2 – âˆš3) is a rational.

Subtract given number form 2, considering 2 is a rational number.

As we know, Difference of two rational numbers is a rational.

So, 2 â€“ (2 â€“ âˆš3 ) is rational

â‡¨ âˆš3 is rational

Which is contradictory.

Thus, (2 â€“ âˆš3) is an irrational.

**(iii) (3 + âˆš2 ) **

Let us assume that (3 + âˆš2 ) is rational.

Subtract 3 form given number, considering 3 is a rational number.

As we know, Difference of two rational numbers is a rational.

(3 + âˆš2 ) â€“ 3 is rational

â‡¨ âˆš2 is rational

Which is contradictory to our assumption.

So, (3 + âˆš2 ) is irrational

**(iv) (2 + âˆš5 )**

Let us assume that (2 + âˆš5 ) is rational.

Subtract 2 form given number, considering 2 is a rational number.

As we know, Difference of two rational numbers is a rational.

(2 + âˆš5) â€“ 2 is rational

â‡¨ âˆš5 is rational

Which is contradictory to our assumption.

So, (2 + âˆš5) is irrational

(**v) (5 + 3âˆš2 )**

Let us assume that (5 + 3âˆš2 ) is rational.

Subtract 5 form given number, considering 5 is a rational number.

As we know, Difference of two rational numbers is a rational.

(5 + 3âˆš2 ) â€“ 5 is rational

â‡¨ 3âˆš2 is rational

And, 3 is rational and âˆš2 is rational.

(Product of two rational numbers is rational)

â‡¨âˆš2 is rational

Which is contradictory to our assumption.

So, (5 + 3âˆš2 ) is irrational

**(vi) 3âˆš7 **

Let us assume that 3âˆš7 is rational

Here 3 is rational and âˆš7 is rational.

As we know, Product of two rational numbers is rational.

But âˆš7 is rational, which is contradictory to our assumption.

So, 3âˆš7 is irrational

**(vii) 3 / âˆš5 **

Let us assume that 3 / âˆš5 is rational

3 / âˆš5 x âˆš5/âˆš5 = 3âˆš5/5

Where 3âˆš5/5 is rational

Which shows that 3/5 is rational and âˆš5 is rational.

But the fact is âˆš5 is an irrational.

Our assumption is wrong, and

3 / âˆš5 is irrational

**(viii)(2 – 3âˆš5)**

Let us assume that 2 – 3âˆš5 is rational.

Subtract given number form 2, considering 2 is a rational number.

As we know, Difference of two rational numbers is a rational.

2 â€“ (2 â€“ 3âˆš5) is rational

â‡¨ 3âˆš5 is rational

Above result is possible if 3 is rational and âˆš5 is rational.

Because, product of two rational numbers is rational

But the fact is âˆš5 is an irrational

Our assumption is wrong, and

(2 â€“ 3âˆš5) is irrational

**(ix) (âˆš3 + âˆš5**)

Let us assume that âˆš3 + âˆš5 is rational

On squaring, we get

(âˆš3 + âˆš5)Â² is rational

â‡¨ 3 + 2âˆš3 x âˆš5 + 5 is rational

â‡¨ 8 + 2âˆš15 is rational

Subtract 8 from above result, considering 8 is a rational number.

As we know, Difference of two rational numbers is a rational.

â‡¨ 8 + 2âˆš15 â€“ 8 is rational

â‡¨ 2âˆš15 is rational

Which is only possible if 2 is rational and âˆš15 is rational.

The fact is âˆš15 is not a rational number.

Our assumption is wrong, and

(âˆš3 + âˆš5) is irrational.

**Question 4. Prove that 1/âˆš3 is irrational.**

**Solution**:

Let us assume that 1/âˆš3 is rational

â‡¨ 1/âˆš3 x âˆš3/ âˆš3 = âˆš3/3 is rational

Which is only possible if 1/3 is rational and âˆš3 is rational. As we know that, Product of two rational numbers is rational

But the fact is, âˆš3 is an irrational.

Which is contradictory to our assumption.

which implies 1/âˆš3 is irrational. Hence proved.

**Question 5: (i) Give an example of two irrationals whose sum is rational.**

**(ii) Give an example of two irrationals whose product is rational.**

**Solution**:

(i) Let us consider two numbers 2 + âˆš3 and 2 â€“ âˆš3 which are irrationals

their sum = (2 + âˆš3) + (2 â€“ âˆš3) = 4; Which is a rational numbers.

(ii) Let us consider two numbers 3 + âˆš2 and 3 â€“ âˆš2 which are irrationals.

their product = (3 + âˆš2) (3 â€“ âˆš2)

= (3)Â² â€“ (âˆš2)Â²

= 9 – 2 = 7; which is a rational number.

**Question 6.**

**State whether the given statement is true or false.**

**(i) The sum of two rationals is always rational.**

**(ii) The product of two rationals is always rational.**

**(iii) The sum of two irrationals is always an irrational.**

**(iv) The product of two irrationals is always an irrational.**

**(v) The sum of a rational and an irrational is irrational.**

**(vi) The product of a rational and an irrational is irrational.**

**Solution**:

(i) True.

(ii) True.

(iii) False.

Reason:

Sum of two irrational can be rational. Let us take an example,

Consider two irrational numbers: (3 + âˆš2) and (3 – âˆš2)

Sum: (3 + âˆš2) + (3 – âˆš2) = 3 + âˆš2 + 3 – âˆš2 = 6 which is rational.

(iv) False.

Reason:

Product of two irrational numbers can be rational. Let us take an example,

Consider two irrational numbers: (3 + âˆš2) and (3 – âˆš2)

Product: (3 + âˆš2)(3 â€“ âˆš2 ) = (3)^{2} â€“ (âˆš2 )^{2} = 9 â€“ 2 = 7; which is rational

(v) True.

(vi) True.

**Question 7. Prove that (2âˆš3 â€“ 1) is an irrational number. **

**Solution**:

Let us assume that (2âˆš3 â€“ 1) is a rational.

Add 1 to the above number, considering 1 is a rational number.

As we know, sum of two rational numbers is a rational.

Sum = 2âˆš3 – 1 + 1 = 2âˆš3

Which is only possible if 2 is rational and âˆš3 is rational.

As we know that product of two rational numbers is rational.

But the fact is âˆš3 is an irrational number which contradicts to our assumption.

So, (2âˆš3 â€“ 1) is an irrational.

**Question 8. Prove that (4 – 5âˆš2 ) is an irrational number.**

**Solution**:

Let us assume that (4 – 5âˆš2 ) is a rational.

Subtract given number form 4, considering 4 is a rational number.

As we know, Difference of two rational numbers is a rational.

4 â€“ (4 â€“ 5âˆš2 ) is rational

â‡¨ 5âˆš2 is rational

Which is only possible if 5 is rational and âˆš2 is rational

As we know, product of two rational number is rational.

But the fact is âˆš2 is an irrational.

Which is contradict to our assumption.

Hence, 4 â€“ 5âˆš2 is irrational. Hence Proved.

**Question 9. Prove that (5 â€“ 2âˆš3) is an irrational number.**

**Solution**:

Let us assume that (5 – 2âˆš3) is a rational.

Subtract given number form 5, considering 5 is a rational number.

As we know, Difference of two rational numbers is a rational.

â‡¨ 5 â€“ (5 â€“ 2âˆš3) is rational

â‡¨ 2âˆš3 is rational

Which is only possible if 2 is rational and âˆš3 is rational

As we know, product of two rational number is rational.

But the fact is âˆš3 is an irrational.

Which is contradict to our assumption.

(5 – 2âˆš3) is an irrational number.

**Question 10: Prove that 5âˆš2 is irrational.**

**Solution**:

Let us assume that 5âˆš2 is a rational.

Which is only possible if 5 is rational and âˆš2 is rational

As we know, product of two rational number is rational.

But the fact is âˆš2 is an irrational.

Which is contradict to our assumption.

5âˆš2 is an irrational. Hence proved.

**Question 11: Prove that 2/âˆš7 is irrational.**

**Solution:**

Let us assume that 2/âˆš7 is a rational number.

2/âˆš7 x âˆš7/âˆš7 = 2âˆš7/7 is a rational number

Which is only possible if 2/7 is rational and âˆš7 is rational.

But the fact is âˆš7 is an irrational.

Which is contradicts to our assumption.

2/âˆš7 is an irrational number. Hence proved.

## RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1D

Class 10 Maths Chapter 1 Real Numbers Exercise 1D, based on the following topics and subtopics:

- Irrational Numbers
- Important results on Irrational