PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Guide

PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Guide

INTRODICTION

The PSEB Solutions for Class 10 Maths Chapter 1, Real Numbers, Exercise 1.3, serve as a comprehensive guide for students. This exercise focuses on important concepts like Euclid’s Division Lemma, the Fundamental Theorem of Arithmetic, and the concept of prime factorization. By solving the questions in this exercise, students can gain a deeper understanding of these topics. The solutions provided in this guide help students grasp the fundamentals of real numbers and strengthen their problem-solving abilities. Additionally, the step-by-step explanations and detailed solutions assist students in learning the correct approach to solving different types of problems. Overall, the PSEB Solutions for Class 10 Maths Chapter 1, Exercise 1.3, are an invaluable resource for students to excel in their mathematics studies.

Question 1. Prove that √5 is irrational.

Solution:

Let us suppose that √5 is rational so we can find integers r and s where s ≠ 0
such that √5 = r/s
Suppose r and s have some common factor other than 1, then divide r and s by the common factor to get :
√5 = a/b where a and b are coprime and b ≠ 0
b√5 = a
Squaring both sides,
⇒ (b√5)² = a²
⇒ b² (√5)² = a²
⇒ 5b² = a² …………..(1)
5 divides a².
By the theorem, if a prime number ‘p’ divides a² then ‘p’ divides a where a is positive integer
⇒ 5 divides a …………(2)
So a = 5c for some integer c.
Put the value of a in (1),
5b² = (5c)²
5b² = 25c²
b² = 5c²
or 5c² = b²
⇒ 5 divides b²
∵ if a prime number ‘p’ divides b², then p divides b ; where b is positive integer.
⇒ 5 divides b ………… (3)
From (2) and (3), a and b have at least 5 as common factor.
But this contradicts the fact that a and b are coprime i.e. no common factor other than 1.
∴ our supposition that √5 is rational wrong.
Hence √5 is irrational.

Question 2. Prove that 3 + 2 √5 is irrational.

Solution:

Let us suppose that 3 + 2√5 is rational.
∴ we can find Co-Prime a and b, where a and b are integers and b ≠ 0
such that 3 + 2√5 = a/b

Since a and b both are integers,

Hence from (1), √5 is rational.
But this contradicts the fact that √5 is irrational.
∴ our supposition is wrong.
Hence 3 + 2√5 is irrational.

Question 3. Prove that the following are irrationals :

(ii) Given that 7√5
Let us suppose that 7^5 is rational
∴ we can find coprime integers a and b where b ≠ 0
such that 7√5 = a/b
⇒ 7b√5 = a
⇒ √5 = a/7b ……………..(1)
Since a, 7 and b are integers, of two integers is a rational number.
i.e., a/7b = rational number
∴ from (1)
√5 = rational number
which contradicts the fact that √5 is irrational number.
∴ Our supposition is wrong.
Hence 7√5 is irrational.

(iii) Given that 6 + √2
Let us suppose that 6 + √2 is rational
∴ we can find coprime integers a and b where b ≠ 0
such that 6 + √2 = a/b

so from (1), √2 = rational number
which contradicts the fact that √2 is irrational number
∴ Our Supposition is wrong.
Hence 6 + √2 is irrational.