RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5
RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5 has been provided here. Students can refer to these questions before their examinations for better preparation.
Ananya Gupta30 Oct, 2024
Share
RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5: In Chapter 1 of RD Sharma Class 10 Maths textbook Exercise 1.5 focuses on the concepts of real numbers and their properties. This exercise includes a variety of problems that help students apply their understanding of rational and irrational numbers, along with their representation on the number line.
Each solution is created to enhance problem-solving skills, providing step-by-step explanations that make it easier for students to grasp the underlying mathematical principles. By working through this exercise, students can strengthen their foundation in real numbers which is important for tackling more advanced topics in mathematics.RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5 Overview
The solutions for RD Sharma's Class 10 Maths Chapter 1 Exercise 1.5 has been created by subject experts at Physics Wallah. These solutions provide clear and comprehensive explanations for each problem, ensuring that students can easily understand the concepts of real numbers and their properties. By breaking down each step, the experts help learners develop a strong foundation in the topic, which is important for mastering more advanced mathematical concepts. These solutions are an excellent resource for students seeking to improve their problem-solving skills and gain confidence in their mathematical abilities.RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5 PDF
The PDF link for RD Sharma Class 10 Maths Chapter 1 Exercise 1.5 is available below. By using this PDF learners can enhance their understanding of real numbers and their properties, facilitating better preparation for exams and assessments.RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5 PDF
RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.5
Here is the RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.5-1. Show that the following numbers are irrational.
(i) 1/√2
Solution:
Consider 1/√2 is a rational number Let us assume 1/√2 = r where r is a rational number On further calculation, we get 1/r = √2 Since r is a rational number, 1/r = √2 is also a rational number But we know that √2 is an irrational number So, our supposition is wrong. Hence, 1/√2 is an irrational number.(ii) 7√5
Solution:
Let’s assume, on the contrary, that 7√5 is a rational number. Then, there exist positive integers a and b such that 7√5 = a/b, where a and b are co-primes ⇒ √5 = a/7b ⇒ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number] This contradicts the fact that √5 is irrational. So, our assumption is incorrect. Hence, 7√5 is an irrational number.(iii) 6 + √2
Solution:
Let’s assume, on the contrary, that 6+√2 is a rational number. Then, there exist coprime positive integers a and b such that 6 + √2 = a/b ⇒ √2 = a/b – 6 ⇒ √2 = (a – 6b)/b ⇒ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 6 + √2 is an irrational number.(iv) 3 − √5
Solution:
Let’s assume, on the contrary, that 3-√5 is a rational number. Then, there exist coprime positive integers a and b such that 3-√5 = a/b ⇒ √5 = a/b + 3 ⇒ √5 = (a + 3b)/b ⇒ √5 is rational [∵ a and b are integers ∴ (a+3b)/b is a rational number] This contradicts the fact that √5 is irrational. So, our assumption is incorrect. Hence, 3-√5 is an irrational number.2. Prove that the following numbers are irrationals.
(i) 2/√7
Solution:
Let’s assume, on the contrary, that 2/√7 is a rational number. Then, there exist coprime positive integers a and b such that 2/√7 = a/b ⇒ √7 = 2b/a ⇒ √7 is rational [∵ 2, a and b are integers ∴ 2b/a is a rational number] This contradicts the fact that √7 is irrational. So, our assumption is incorrect. Hence, 2/√7 is an irrational number.(ii) 3/(2√5)
Solution:
Let’s assume, on the contrary, that 3/(2√5) is a rational number. Then, there exist coprime positive integers a and b such that 3/(2√5) = a/b ⇒ √5 = 3b/2a ⇒ √5 is rational [∵ 3, 2, a and b are integers ∴ 3b/2a is a rational number] This contradicts the fact that √5 is irrational. So, our assumption is incorrect. Hence, 3/(2√5) is an irrational number.(iii) 4 + √2
Solution:
Let’s assume, on the contrary, that 4 + √2 is a rational number. Then, there exist coprime positive integers a and b such that 4 + √2 = a/b ⇒ √2 = a/b – 4 ⇒ √2 = (a – 4b)/b ⇒ √2 is rational [∵ a and b are integers ∴ (a – 4b)/b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 4 + √2 is an irrational number.(iv) 5√2
Solution:
Let’s assume, on the contrary, that 5√2 is a rational number. Then, there exist positive integers a and b such that 5√2 = a/b, where a and b are co-primes ⇒ √2 = a/5b ⇒ √2 is rational [∵ a and b are integers ∴ a/5b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 5√2 is an irrational number.3. Show that 2 − √3 is an irrational number.
Solution:
Let’s assume, on the contrary, that 2 – √3 is a rational number. Then, there exist coprime positive integers a and b such that 2 – √3= a/b ⇒ √3 = 2 – a/b ⇒ √3 = (2b – a)/b ⇒ √3 is rational [∵ a and b are integers ∴ (2b – a)/b is a rational number] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 2 – √3 is an irrational number.4. Show that 3 + √2 is an irrational number .
Solution:
Let’s assume, on the contrary, that 3 + √2 is a rational number. Then, there exist coprime positive integers a and b such that 3 + √2= a/b ⇒ √2 = a/b – 3 ⇒ √2 = (a – 3b)/b ⇒ √2 is rational [∵ a and b are integers ∴ (a – 3b)/b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 3 + √2 is an irrational number.5. Prove that 4 − 5√2 is an irrational number.
Solution:
Let’s assume, on the contrary, that 4 – 5√2 is a rational number. Then, there exist coprime positive integers a and b such that 4 – 5√2 = a/b ⇒ 5√2 = 4 – a/b ⇒ √2 = (4b – a)/(5b) ⇒ √2 is rational [∵ 5, a and b are integers ∴ (4b – a)/5b is a rational number] This contradicts the fact that √2 is irrational. So, our assumption is incorrect. Hence, 4 – 5√2 is an irrational number.6. Show that 5 − 2√3 is an irrational number.
Solution:
Let’s assume, on the contrary, that 5 – 2√3 is a rational number. Then, there exist coprime positive integers a and b such that 5 – 2√3 = a/b ⇒ 2√3 = 5 – a/b ⇒ √3 = (5b – a)/(2b) ⇒ √3 is rational [∵ 2, a and b are integers ∴ (5b – a)/2b is a rational number] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 5 – 2√3 is an irrational number.7. Prove that 2√3 − 1 is an irrational number.
Solution:
Let’s assume, on the contrary, that 2√3 – 1 is a rational number. Then, there exist coprime positive integers a and b such that 2√3 – 1 = a/b ⇒ 2√3 = a/b + 1 ⇒ √3 = (a + b)/(2b) ⇒ √3 is rational [∵ 2, a and b are integers ∴ (a + b)/2b is a rational number] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, 2√3 – 1 is an irrational number.8. Prove that 2 − 3√5 is an irrational number.
Solution:
Let’s assume, on the contrary, that 2 – 3√5 is a rational number. Then, there exist co-prime positive integers a and b such that 2 – 3√5 = a/b ⇒ 3√5 = 2 – a/b ⇒ √5 = (2b – a)/(3b) ⇒ √5 is rational [∵ 3, a and b are integers ∴ (2b – a)/3b is a rational number] This contradicts the fact that √5 is irrational. So, our assumption is incorrect. Hence, 2 – 3√5 is an irrational number.9. Prove that √5 + √3 is irrational.
Solution:
Let’s assume, on the contrary, that √5 + √3 is a rational number. Then, there exist coprime positive integers a and b such that √5 + √3 = a/b ⇒ √5 = (a/b) – √3 ⇒ (√5) 2 = ((a/b) – √3) 2 [Squaring on both sides] ⇒ 5 = (a 2 /b 2 ) + 3 – (2√3a/b) ⇒ (a 2 /b 2 ) – 2 = (2√3a/b) ⇒ (a/b) – (2b/a) = 2√3 ⇒ (a 2 – 2b 2 )/2ab = √3 ⇒ √3 is rational [∵ a and b are integers ∴ (a 2 – 2b 2 )/2ab is rational] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, √5 + √3 is an irrational number.10. Prove that √2 + √3 is irrational.
Solution:
Let’s assume, on the contrary, that √2 + √3 is a rational number. Then, there exist coprime positive integers a and b such that √2 + √3 = a/b ⇒ √2 = (a/b) – √3 ⇒ (√2) 2 = ((a/b) – √3) 2 [Squaring on both sides] ⇒ 2 = (a 2 /b 2 ) + 3 – (2√3a/b) ⇒ (a 2 /b 2 ) + 1 = (2√3a/b) ⇒ (a/b) + (b/a) = 2√3 ⇒ (a 2 + b 2 )/2ab = √3 ⇒ √3 is rational [∵ a and b are integers ∴ (a 2 + 2b 2 )/2ab is rational] This contradicts the fact that √3 is irrational. So, our assumption is incorrect. Hence, √2 + √3 is an irrational number.11. Prove that for any prime positive integer p, √p is an irrational number.
Solution:
Consider √p as a rational number Assume √p = a/b where a and b are integers and b ≠ 0 By squaring on both sides p = a 2 /b 2 pb = a 2 /b p and b are integers pb= a 2 /b will also be an integer But we know that a 2 /b is a rational number, so our supposition is wrong Therefore, √p is an irrational number.12. If p and q are prime positive integers, prove that √p + √q is an irrational number.
Solution:
Let’s assume, on the contrary, that √p + √q is a rational number. Then, there exist coprime positive integers a and b such that √p + √q = a/b ⇒ √p = (a/b) – √q ⇒ (√p) 2 = ((a/b) – √q) 2 [Squaring on both sides] ⇒ p = (a 2 /b 2 ) + q – (2√q a/b) ⇒ (a 2 /b 2 ) – (p+q) = (2√q a/b) ⇒ (a/b) – ((p+q)b/a) = 2√q ⇒ (a 2 – b 2 (p+q))/2ab = √q ⇒ √q is rational [∵ a and b are integers ∴ (a 2 – b 2 (p+q))/2ab is rational] This contradicts the fact that √q is irrational. So, our assumption is incorrect. Hence, √p + √q is an irrational number.Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5
- Concept Clarity : The solutions provided for Exercise 1.5 help students grasp fundamental concepts of real numbers, including their properties and applications. This clarity is important for building a strong foundation in mathematics.
- Practice and Reinforcement : Regularly solving these exercises reinforces learning and aids retention. It provides students with ample practice to master the topic which is important for success in exams.
- Exam Preparation : The solutions are a valuable resource for exam preparation. Students can use them to familiarize themselves with the types of questions that may appear on tests and to practice under similar conditions.
- Self-Assessment : By attempting the exercises and then checking their answers against the solutions, students can assess their understanding of the material. This self-evaluation helps identify areas that need further attention or improvement.
- Confidence Building : Solving problems boosts students confidence in their mathematical abilities. With increased confidence, they are more likely to approach complex problems and challenges in future chapters.
RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5 FAQs
What are real numbers?
Real numbers include all the numbers on the number line. This includes rational numbers (like fractions and integers) and irrational numbers (like the square root of 2 and π).
Are all integers real numbers?
Yes, all integers are considered real numbers. This includes positive integers, negative integers and zero.
Can real numbers be negative?
Yes, real numbers can be positive, negative, or zero. The set of real numbers encompasses both the negative and positive sides of the number line.
How are real numbers represented on the number line?
Real numbers are represented as points on a continuous line where every point corresponds to a unique real number.
🔥 Trending Blogs
Talk to a counsellorHave doubts? Our support team will be happy to assist you!
Check out these Related Articles
Join 15 Million students on the app today!
Live & recorded classes available at easeDashboard for progress trackingMillions of practice questions at your fingertips
Free Learning Resources
PW Books
Notes (Class 10-12)
PW Study Materials
Notes (Class 6-9)
Ncert Solutions
Govt Exams
Class 6th to 12th Online Courses
Govt Job Exams Courses
UPSC Coaching
Defence Exam Coaching
Gate Exam Coaching
Other Exams
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Copyright © 2025 Physicswallah Limited All rights reserved.