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RD Sharma 10th class Chapter 1- Updated 2022-2023

Class 10
RD Sharma 10th class Chapter 1- Updated 2022-2023

In RD Sharma's 10th class Chapter 1 -Real Numbers introduces students to the fundamental concept of real numbers and their properties. This chapter lays the foundation for several mathematical concepts and theories that students will encounter in higher classes. Through this chapter, students will not only learn about the properties of real numbers but also understand the relationships between different types of numbers. In this article, we will explore the key concepts and topics covered in RD Sharma's 10th class Chapter 1- Real Number.

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RD Sharma 10th class Chapter 1-  Real Number 

RD Sharma 10th class Chapter 1- Real Numbers Exercise 1.1 Page No: 1.10

1. If a and b are two odd positive integers such that a > b, then prove that one of the two numbers (a+b)/2 and (a-b)/2 is odd, and the other is even.

Solution:

We know that any odd positive integer is of form 4q+1 or 4q+3 for some whole number q.

Now that it’s given a > b

So, we can choose a= 4q+3 and b= 4q+1.

∴ (a+b)/2 = [(4q+3) + (4q+1)]/2

⇒ (a+b)/2 = (8q+4)/2

⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number.

Now, doing (a-b)/2

⇒ (a-b)/2 = [(4q+3)-(4q+1)]/2

⇒ (a-b)/2 = (4q+3-4q-1)/2

⇒ (a-b)/2 = (2)/2

⇒ (a-b)/2 = 1, which is an odd number.

Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd, and the other is even.

2. Prove that the product of two consecutive positive integers is divisible by 2.

RD Sharma 10th class Chapter 1- Real Numbers

Solution:

Let’s consider two consecutive positive integers as (n-1) and n.

∴ Their product = (n-1) n

= n– n

We know that any positive integer is of form 2q or 2q+1. (From Euclid’s division lemma for b= 2)

So, when n= 2q

We have,

⇒ n– n = (2q)– 2q

⇒ n– n = 4q-2q

⇒ n– n = 2(2q-q)

Thus, n– n is divisible by 2.

Now, when n= 2q+1

We have,

⇒ n– n = (2q+1)2 – (2q-1)

⇒ n– n = (4q2+4q+1 – 2q+1)

⇒ n– n = (4q2+2q+2)

⇒ n– n = 2(2q2+q+1)

Thus, n– n is divisible by 2 again.

Hence, the product of two consecutive positive integers is divisible by 2.

3. Prove that the product of three consecutive positive integers is divisible by 6.

RD Sharma 10th class Chapter 1- Real Numbers

Solution:

Let n be any positive integer.

Thus, the three consecutive positive integers are n, n+1 and n+2.

We know that any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)

So,

For n= 6q,

⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2)

⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]

For n= 6q+1,

⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)

⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)]

For n= 6q+2,

⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)

⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)]

For n= 6q+3,

⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)

⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]

For n= 6q+4,

⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)

⇒ n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+2)(3q+1)(2q+2)]

For n= 6q+5,

⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)

⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+5)(q+1)(6q+7)]

Hence, the product of three consecutive positive integers is divisible by 6.

4. For any positive integer n, prove that n3 – n is divisible by 6.

Solution:

Let n be any positive integer. And since any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)

We have n3 – n = n(n2-1)= (n-1)n(n+1),

For n= 6q,

⇒ (n-1)n(n+1)= (6q-1)(6q)(6q+1)

⇒ (n-1)n(n+1)= 6[(6q-1)q(6q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q-1)q(6q+1)]

For n= 6q+1,

⇒ (n-1)n(n+1)= (6q)(6q+1)(6q+2)

⇒ (n-1)n(n+1)= 6[q(6q+1)(6q+2)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]

For n= 6q+2,

⇒ (n-1)n(n+1)= (6q+1)(6q+2)(6q+3)

⇒ (n-1)n(n+1)= 6[(6q+1)(3q+1)(2q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+1)

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RD Sharma 10th class Chapter 1- Updated 2022-2023 

 

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