RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1: Natural numbers, integers, rational and irrational numbers are all elements of the universal set of real numbers. We will learn about integer divisibility as well as several key features of integers in RD Sharma Solutions Class 10 Exercise 1.1. In this exercise, the main focus is Euclid’s division Lemma. Students can use the RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1 PDF provided below to solve these questions effectively.

Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1 Free PDF

RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1 PDF

Access answers to RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1- Important Question with Answers

Question 1.
If a and b are two odd positive integers such that a > b, then prove that one of the two numbers a+b2 and a−b2 are odd and the other is even.
Solution:
a and b are two odd numbers such that a > b
Let a = 2n + 1, then b = 2n + 3

Question 2.
Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let n and n + 1 are two consecutive positive integer
We know that n is of the form n = 2q and n + 1 = 2q + 1
n (n + 1) = 2q (2q + 1) = 2 (2q² + q)
Which is divisible by 2
If n = 2q + 1, then
n (n + 1) = (2q + 1) (2q + 2)
= (2q + 1) x 2(q + 1)
= 2(2q + 1)(q + 1)
Which is also divisible by 2
Hence the product of two consecutive positive integers is divisible by 2.

Question 3.
Prove that the product of three consecutive positive integers is divisible by 6.
Solution:
Let n be the positive any integer Then
n(n + 1) (n + 2) = (n² + n) (n + 2)

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

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


Which is divisible by 6
Hence we can similarly, prove that n² – n is divisible by 6 for any positive integer n.
Hence proved.

Question 5.
Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.
Solution:
Let n = 6q + 5, where q is a positive integer
We know that any positive integer is of the form 3k or 3k + 1 or 3k + 2, 1
q = 3k or 3k + 1 or 3k + 2
If q = 3k, then n = 6q + 5

Question 6.
Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Solution:
Let a be any positive integer
Then a = 5m + 1
a² = (5m + 1 )² = 25m² + 10m + 1
= 5 (5m² + 2m) + 1
= 5q + 1 where q = 5m² + 2m
Which is of the same form as given
Hence proved.

Question 7.
Prove that the square of any positive, integer is of the form 3m or, 3m + 1 but not of form 3m + 2.
Solution:
Let a be any positive integer
Let it be in the form of 3m or 3m + 1
Let a = 3q, then

Hence proved.

Question 8.
Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
Solution:
Let a be the positive integer and
Let a = 4m

Hence proved.

Question 9.
Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
Solution:
Let a be the positive integer, and
Let a = 5m, then

Question 10.
Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Solution:
Let n is any positive odd integer
Let n = 4p + 1, then
(4p + 1)² = 16p² + 8p + 1
n² = 8p (2p + 1) + 1
= 8q + 1 where q = p(2p + 1)
Hence proved.

Question 11.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let n be any positive odd integer and
let n = 6q + r
=> 6q + r, b = 6, and 0 ≤ r < 6
or r = 0, 1, 2, 3, 4, 5
If n = 6q = 2 x 3q
But it is not odd
When n = 6q + 1 which is odd
When n = 6q + 2 which is not odd = 2 (3q+ 1)
When n = 6q + 3 which is odd
When n = 6q + 4 = 2 (3q + 2) which is not odd
When n = 6q + 5, which is odd
Hence 6q + 1 or 6q + 3 or 6q + 5 are odd numbers.

Question 12.
Show that the square of any positive integer cannot be of form 6m + 2 or 6m + 5 for any integer m. [NCERT Exemplar]
Solution:
Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that


Hence, the square of any positive integer cannot be of form 6m + 2 or 6m + 5 for any integer m.

Question 13.
Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5. [NCERT Exemplar]
Solution:
Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that



Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the forms 6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

Question 14.
Show that one and only one out of n, n + 4, n + 8, n + 12, and n + 16 is divisible by 5, where n is any positive integer. [NCERT Exemplar]
Solution:
Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer.
Then, let n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4 for q ∈N [By Euclid’s algorithm]
Then, in each case, if we put the different values of n in the given numbers. We definitely get one and only one of the given numbers is divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12, and n + 16 is divisible by 5.
Alternate Method
On dividing on n by 5, let q be the quotient and r be the remainder.
Then n = 5q + r, where 0 ≤ r < 5. n = 5q + r, where r = 0, 1, 2, 3, 4
=> n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4
Case I: If n = 5q, then n is only divisible by 5. .
Case II: If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1), which is only divisible by 5.
So, in this case, (n + 4) is divisible by 5.
Case III : If n = 5q + 3, then n + 2 = 5q + 3 + 12 = 5q + 15 = 5(q + 3), which is divisible by 5.
So, in this case (n + 12) is only divisible by 5.
Case IV : If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4), which is divisible by 5.
So, in this case, (n + 16) is only divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12, and n + 16 is divisible by 5, where n is any positive integer.

Question 15.
Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer? [NCERT Exemplar]
Solution:
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of form 6m + 1, 6m + 3, or 6m + 5 Thus we have:

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Question 16.
A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m, or 3m + 2 for some integer m? Justify your answer.
Solution:
No, by Euclid’s Lemma, b = aq + r, 0 ≤ r < a
Here, b is any positive integer

Question 17.
Show that the square of any positive integer cannot be of form 3m + 2, where m is a natural number.
Solution:
By Euclid’s lemma, b = aq + r, 0 ≤ r ≤ a
Here, b is any positive integer,
a = 3, b = 3q + r for 0 ≤ r ≤ 2
So, any positive integer is of the form 3k, 3k + 1 or 3k + 2

Which is in the form of 3m + 1. Hence, a square of any positive number cannot be of form 3m + 2.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.1