In this chapter, we provide RD Sharma Solutions for Class 10 Solutions Chapter 1 Real Numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 10 Solutions Chapter 1 Real Numbers pdf, free RD Sharma Solutions for Class 10 Solutions Chapter 1 Real Numbers book pdf download. Now you will get step by step solution to each question.

## RD Sharma Class 10 Solutions Chapter 1 Real Numbers

### Class 10 RD Sharma Solutions Chapter 1 Real Numbers Ex 1.1

Question 1.

If a and b are two odd positive integers such that a > b, then prove that one of the two numbers (frac { a+b }{ 2 }) and (frac { a-b }{ 2 }) is 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 (2q^{2} + 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 integer is divisible by 6.

Solution:

Let n be the positive any integer Then

n(n + 1) (n + 2) = (n^{2} + 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 n^{3} – n is divisible by 6.

Solution:

Which is divisible by 6

Hence we can similarly, prove that n^{2} – 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

a^{2} = (5m + 1 )^{2} = 25m^{2} + 10m + 1

= 5 (5m^{2} + 2m) + 1

= 5q + 1 where q = 5m^{2} + 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 the 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} = 16p^{2} + 8p + 1

n^{2} = 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 the 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 the 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 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 the 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 the 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, square of any positive number cannot be of the form 3m + 2.

### RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Solution 11:

**All Chapter RD Sharma Solutions For Class 10 Maths**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.