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NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

Edited By Ramraj Saini | Updated on Nov 02, 2023 01:21 PM IST | #CBSE Class 10th

NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

This article discuss NCERT Solutions for class 10 maths ex 1.3 with all questions and answer in step by step manner. These NCERT solutions for exercise are created by expert team at Careers360 keeping in mind the need of an average students and latest syllabus of CBSE 20023-24. This exercise deals with irrational numbers which are real numbers that cannot be written in simple fractions for example π, √2, √3, √5, √7 - 3, etc. One of the theorems we use in this exercise is the Fundamental Theorem of Arithmetic.

This Story also Contains
  1. NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers
  2. Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 Real Numbers
  3. Access Exercise 1.3 Class 10 Maths Answers
  4. Benefits of NCERT Solutions for Class 10 Maths Exercise 1.3
  5. NCERT Solutions of Class 10 Subject Wise
  6. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers
NCERT Solutions for Exercise 1.3 Class 10 Maths Chapter 1 - Real Numbers

There are four exercise in this chapter which are listed below. interested students can refer them and practice problems to command the concepts.

Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3 Real Numbers

Download PDF

Access Exercise 1.3 Class 10 Maths Answers

Real Numbers Class 10 Chapter 1- Exercise 1.3

Q1 Prove that \sqrt 5 is irrational.

Answer:

Let us assume \sqrt{5} is rational.

It means \sqrt{5} can be written in the form \frac{p}{q} where p and q are co-primes and q\neq 0

\\\sqrt{5}=\frac{p}{q}

Squaring both sides we obtain

\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\ 5=\frac{p^{2}}{q^{2}}\\ p^{2}=5q^{2}

From the above equation, we can see that p 2 is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number. (i)

Therefore p can be written as 5r

p = 5r

p 2 = (5r) 2

5q 2 = 25r 2

q 2 = 5r 2

From the above equation, we can see that q 2 is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number. (ii)

From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that \sqrt{5} is rational was wrong. Hence proved that \sqrt{5} is irrational.

Q2 Prove that 3 + 2 \sqrt 5 is irrational.

Answer:

Let us assume 3 + 2 \sqrt 5 is rational.

This means 3 + 2 \sqrt 5 can be wriiten in the form \frac{p}{q} where p and q are co-prime integers.

\\3+2\sqrt{5}=\frac{p}{q}\\ 2\sqrt{5}=\frac{p}{q}-3\\ \sqrt{5}=\frac{p-3q}{2q}\\

As p and q are integers \frac{p-3q}{2q}\\ would be rational, this contradicts the fact that \sqrt{5} is irrational. This contradiction arises because our initial assumption that 3 + 2 \sqrt 5 is rational was wrong. Therefore 3 + 2 \sqrt 5 is irrational.

Q3 Prove that the following are irrationals :

(i) 1/ \sqrt 2

Answer:

Let us assume \frac{1}{\sqrt{2}} is rational.

This means \frac{1}{\sqrt{2}} can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\\frac{1}{\sqrt{2}}=\frac{p}{q}\\ \sqrt{2}=\frac{q}{p}

Since p and q are co-prime integers \frac{q}{p} will be rational, this contradicts the fact that \sqrt{2} is irrational. This contradiction arises because our initial assumption that \frac{1}{\sqrt{2}} is rational was wrong. Therefore \frac{1}{\sqrt{2}} is irrational.

Q3 (2) Prove that the following are irrationals :

(ii) 7 \sqrt 5

Answer:

Let us assume 7 \sqrt 5 is rational.

This means 7 \sqrt 5 can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\7\sqrt{5}=\frac{p}{q}\\ \sqrt{5}=\frac{p}{7q}

As p and q are integers \frac{p}{7q}\\ would be rational, this contradicts the fact that \sqrt{5} is irrational. This contradiction arises because our initial assumption that 7 \sqrt 5 is rational was wrong. Therefore 7 \sqrt 5 is irrational.

Q3 (3) Prove that the following are irrationals : 6 + \sqrt 2

Answer:

Let us assume 6 + \sqrt 2 is rational.

This means 6 + \sqrt 2 can be written in the form \frac{p}{q} where p and q are co-prime integers.

\\6+\sqrt{2}=\frac{p}{q}\\ \sqrt{2}=\frac{p}{q}-6\\ \sqrt{2}=\frac{p-6q}{q}

As p and q are integers \frac{p-6q}{q} would be rational, this contradicts the fact that \sqrt{2} is irrational. This contradiction arises because our initial assumption that 6 + \sqrt 2 is rational was wrong. Therefore 6 + \sqrt 2 is irrational.

Benefits of NCERT Solutions for Class 10 Maths Exercise 1.3

  • NCERT solutions for Class 10 Maths is considered the best material for 10th class maths exercise 1.3 answers.
  • NCERT Class 10th Maths chapter 1 exercise 1.3, contains all important questions from exam point of view.
  • Exercise 1.3 Class 10 Maths, is based on irrational numbers and based on the Fundamental Theorem of Arithmetic, which are important concepts of the chapter. students can also study Real Numbers Class 10 Notes to revise the concepts quickly.
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sai raja 01 September,2024
show the previous year exams ?

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best of luck!

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