NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.3 is an essential section of Chapter 1 for understanding irrational numbers and their properties. This exercise provides a clear description of irrational numbers, enabling students to reinforce their knowledge of these concepts. In line with the new CBSE syllabus and examination guidelines, it is important to understand the significance of irrational numbers in the entire number system. Let's discuss this significant topic in Class 10th in greater detail.

1.0Download NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 : Free PDF

2.0Introduction to Irrational Numbers

Irrational numbers can be defined as numbers that cannot be written as a fraction of two integers that are in the form $\frac{p}{q}$, where p and q are integers, and $q\ne 0$. These numbers have non-terminating and non-repeating decimal expansions. Some examples of these numbers are $\sqrt{2},\sqrt{3},\pi ,e$ (Euler's number), etc.

3.0Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 Overview: Key Concepts

Decimal Expansion of Rational Numbers:

1. Non-terminating Decimal Expansion: A decimal expansion is referred to as non-terminating if the digits following the decimal point never stop but go on forever. For instance, the decimal expansion of is 3.14159..., and it just keeps going without a stoppage.
2. Non-repeating Decimal Expansion: A decimal expansion is non-repeating if the numbers following the decimal point never consist of a repeating sequence or pattern. For instance, the decimal expansion is 3.14159... does not contain any repeating pattern, and hence, it is both non-terminating and non-repeating.

Evidence that 2 is Irrational Number: 

We use the method contradiction to prove the irrationality of the number 2. The method includes the contradictory assumption that 2 is a rational number. When proving it, the number cannot be written as a ratio of two integers, and so it is considered an irrational number.

Prime Factorisation and Divisibility: 

The theorem employed in the proof of 2 irrationality relies on the fact that whenever a prime factorises a square, it would also divide the original number. This is one of the major results in the field of number theory. According to this theory, if a prime number say p is given and it divides the square of a number say a², then p must also divide a. Here “a” is a positive integer. 

Review of Irrational Numbers: 

The most important aspect of irrational numbers is that they cannot be expressed as a fraction of integers and possess decimal expansions that do not terminate or repeat. In the exercise, we will be proving the irrationality of some of these numbers. 

Apart from these numbers, the exercise will also include proving compounds of rational and irrational numbers. To prove this we follow some rules of simple arithmetic that are: 

• Rational and Irrational Combined: The addition or subtraction of an irrational and a rational number always results in an irrational number. This is true because a rational number can always be written in the form of a fraction, whereas an irrational number cannot. When added or subtracted together, the solution cannot be resolved into a rational form, resulting in an irrational product.
• Product of Rational and Irrational Numbers: The result or product of a non-zero rational number and an irrational number is also irrational. When you divide or multiply a rational number by an irrational number, the number remains irrational because its result cannot be represented as a fraction.

The concepts mentioned above are crucial for understanding the concepts of rational and irrational numbers. So, begin your journey from here by mastering NCERT Solutions Class 10 Maths Chapter 1 - Exercise 1.3.

4.0NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.3: Detailed Solutions

1. Prove that √5 is irrational.

Solution:

Let us assume, to the contrary, that √5 is rational.

So, we can find coprime integers a and b (b ≠ 0) such that √5 = a/b, b ≠ 0, a, b ∈ I.

√5 b = a

Squaring on both sides, we get 5b² = a².

Therefore, 5 divides a².

(by fundamental theorem of arithmetic) Therefore, 5 divides a.

So, we can write a = 5c for some integer c.

Substituting for a, we get 5b² = 25c² ⇒ b² = 5c².

This means that 5 divides b², and so 5 divides b.

Therefore, a and b have at least 5 as a common factor.

But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arisen because of our incorrect assumption that √5 is rational.

So, we conclude that √5 is irrational.

2. Prove that 3 + 2√5 is irrational.

Solution:

Let us assume, to the contrary, that 3 + 2√5 is rational. So, we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b, b ≠ 0, a, b ∈ I.

Therefore, a/b - 3 = 2√5 ⇒ (a - 3b)/b = 2√5 ⇒ (a - 3b)/(2b) = √5.

Since a and b are integers, we get (a - 3b)/(2b) is rational, and so √5 is rational.

But this contradicts the fact that √5 is irrational. This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational.

So, we conclude that 3 + 2√5 is irrational.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Solution:

(i) 1/√2

Let us assume, to the contrary, that 1/√2 is rational.

That is, we can find coprime integers p and q (q ≠ 0) such that 1/√2 = p/q, where p, q ∈ I.

Therefore, q/p = √2.

Since p and q are integers, q/p is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

So, we conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume, to the contrary, that 7√5 is rational.

That is, we can find coprime integers a and b (b ≠ 0) such that 7√5 = a/b, where a, b ∈ I.

Therefore, a/(7b) = √5.

Since a and b are integers, a/(7b) is rational, and so √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 7√5 is rational.

So, we conclude that 7√5 is irrational.

(iii) 6 + √2

Let us assume, to the contrary, that 6 + √2 is rational.

That is, we can find coprime integers a and b (b ≠ 0) such that 6 + √2 = a/b, where a, b ∈ I.

Therefore, a/b - 6 = √2, which implies (a - 6b)/b = √2.

Since a and b are integers, (a - 6b)/b is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 6 + √2 is rational.

So, we conclude that 6 + √2 is irrational.

5.0Benefits of NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

• Covers important questions frequently asked in CBSE board exams.
• Strengthens basics for advanced topics in algebra and number theory.
• Helps prepare for competitive exams like NTSE, and Olympiads.