Class 10 RD Sharma Solutions - Chapter 1 Real Numbers - Exercise 1.1 | Set 1

In Class 10 Mathematics, Chapter 1 of RD Sharma's textbook deals with Real Numbers. This chapter lays the foundation for understanding the different types of numbers and their properties. Exercise 1.1, Set 1 focuses on fundamental concepts related to the real numbers including their classification and the basic operations performed on them. This exercise helps students solidify their understanding of the number system which is crucial for solving more complex mathematical problems.

Real Numbers

The Real numbers encompass all the numbers that can be found on the number line. They include:

• Natural Numbers: The Counting numbers starting from 1, 2, 3, and so on.
• Whole Numbers: The Natural numbers including 0 (0, 1, 2, 3, ...).
• Integers: Whole numbers and their negative counterparts (-1, -2, 0, 1, 2, ...).
• Rational Numbers: The Numbers that can be expressed as the ratio of two integers where the denominator is not zero.
• Irrational Numbers: The Numbers that cannot be expressed as a simple fraction and their decimal representation is non-terminating and non-repeating

Question 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:

Any odd positive integer is of the form 2q+1 or, 2q+3 for some whole number q.

a > b (given)

a = 2q+3 and b = 2q+1.

Therefore, (a+b)/2 = [(2q+3) + (2q+1)]/2

⇒ (a+b)/2 = (4q+4)/2

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

Now, (a-b)/2

⇒ (a-b)/2 = [(2q+3)-(2q+1)]/2

⇒ (a-b)/2 = (2q+3-2q-1)/2

⇒ (a-b)/2 = (2)/2

⇒ (a-b)/2 = 1 which is an odd number.

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

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

Solution:

Let two consecutive positive integers as n and n+1

Therefore, product = n(n+1)

= n² + n

Any positive integer is of the form 2q or 2q+1. 

Let n = 2q

⇒ n² + n = (2q)² +2q

⇒n² + n = 4q² +2q

⇒ n² + n = 2(2q² +q)

Therefore, n² + n is divisible by 2.

n = 2q+1

⇒ n² + n = (2q+1)² + (2q+1)

⇒ n² + n = (4q²+4q+1 +2q+1)

⇒ n² + n = (4q²+6q+2)

⇒ n² + n = 2(2q²+3q+1)

Thus, n² + n is divisible by 2 

Therefore, 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 any positive integer.

Three consecutive positive integers are n, n+1 and n+2.

Any positive integer can be of the form 6q or 6q+1 or 6q+2 or 6q+3 or 6q+4 or 6q+5. 

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.

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

Solution:

Let, n be any positive integer. Any positive integer can be of the form 6q,6q+1, 6q+2,6q+3,6q+4,6q+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)(3q+1)(2q+1)]

For n= 6q+3,

⇒ (n-1)n(n+1)= (6q+2)(6q+3)(6q+4)

⇒ (n-1)n(n+1)= 6[(3q+1)(2q+1)(6q+4)]

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

For n= 6q+4,

⇒ (n-1)n(n+1)= (6q+3)(6q+4)(6q+5)

⇒ (n-1)n(n+1)= 6[(2q+1)(3q+2)(6q+5)]

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

For n= 6q+5,

⇒ (n-1)n(n+1)= (6q+4)(6q+5)(6q+6)

⇒ (n-1)n(n+1)= 6[(6q+4)(6q+5)(q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+4)(6q+5)(q+1)]

Hence, for any positive integer n, n³ – n is divisible by 6.

Question 5.  Prove that if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.

Solution:

Let n= 6q+5 

Any positive integer can be of the form 3k,3k+1,3k+2.

Therefore, q can be 3k,3k+1,3k+2.

If q= 3k, then

⇒ n= 6q+5

⇒ n= 6(3k)+5

⇒ n= 18k+5 = (18k+3)+ 2

⇒ n= 3(6k+1)+2

Therefore, n= 3m+2, where m is 6k+1

If q= 3k+1, then

⇒ n= 6q+5

⇒ n= 6(3k+1)+5

⇒ n= 18k+6+5 = (18k+9)+ 2

⇒ n= 3(6k+3)+2

Therefore, n= 3m+2, where m is 6k+3

If q= 3k+2, then

⇒ n= 6q+5

⇒ n= 6(3k+2)+5

⇒ n= 18k+12+5 = (18k+15)+ 2

⇒ n= 3(6k+5)+2

⇒ n= 3m+2, where m is 6k+5

Therefore, if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely,

Let n= 3q+2

And we know that a positive integer can be of the form 6k,6k+1,6k+2,6k+3,6k+4,6k+5.

So, now if q=6k+1 then

⇒ n= 3q+2

⇒ n= 3(6k+1)+2

⇒ n= 18k + 5

⇒ n= 6(3k)+5

⇒ n=6m+5

m is some integer

q=6k+2 then

⇒ n= 3q+2

⇒ n= 3(6k+2)+2

⇒ n= 18k + 6 +2 = 18k+8

⇒ n= 6 (3k + 1) + 2

⇒ n = 6m + 2

⇒ n= 6m+2, 

m is some integer

It is not of the form 6q + 5.

Therefore, if n is of the form 3q + 2, then is necessary won’t be of the form 6q + 5.

Question 6.  Prove that square of any positive integer of the form 5q + 1 is of the same form.

Solution:

n=5q+1

On squaring it,

⇒ n²= (5q+1)²

⇒ n²= (25q²+10q+1)

⇒ n²= 5(5q²+2q)+1

⇒ n²= 5m+1, where m is some integer. [For m = 5q²+2q]

Therefore, the square of any positive integer of the form 5q + 1 is of the same form.

Question 7. Prove that the square of any positive integer is of form 3m or 3m + 1 but not of form 3m + 2.

Solution:

Let positive integer be of the form 3q, 3q+1,3q+2. (From Euclid’s division lemma for b= 3)

If n= 3q,

Then, on squaring

⇒ n²= (3q)² = 9q²

⇒ n²= 3(3q²)

⇒ n²= 3m, where m is some integer [m = 3q²]

If n= 3q+1,

Then, on squaring

⇒ n²= (3q+1)² = 9q² + 6q + 1

⇒ n²= 3(3q² +2q) + 1

⇒ n²= 3m + 1, where m is some integer [m = 3q² +2q]

If n= 3q+2,

Then, on squaring

⇒ n²= (3q+2)² = 9q² + 12q + 4

⇒ n²= 3(3q² + 4q + 1) + 1

⇒ n²= 3m + 1, where m is some integer [m = 3q² + 4q + 1]

Therefore, square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Question 8.  Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

Solution:

a=bq+r (by Euclid's division lemma)

According to the question, b = 4.

a = 4p + r, 0 < r < 4

r = 0, a = 4p 

a² = 16p² = 4(4p²) = 4q, where q = 4p²

r = 1, a = 4p + 1

a² = (4p + 1)² = 16p² + 1 + 8p = 4(4p + 2) + 1 = 4q + 1, where q = (4p + 2)

r = 2, a = 4p + 2

a² = (4p + 2)² = 16p² + 4 + 16p = 4(4p² + 4p + 1) = 4q, where q = 4p² + 4p + 1

r = 3, a = 4k + 3

a² = (4p + 3)² = 16p² + 9 + 24p = 4(4p² + 6p + 2) + 1

= 4q + 1, where q = 4p² + 6p + 2

Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Question 9. Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Solution:

According to Euclid’s division lemma,

a = bm+r

According to the question, b = 5.

a = 5m + r, 0 < r < 5

r = 0 a = 5m 

a² = 25m² = 5(5m²) = 5q, where q = 5m²

When r = 1, we get, a = 5m + 1

a² = (5m + 1)² = 25m² + 1 + 10m = 5m(5m + 2) + 1 = 5q + 1, where q = m(5m + 2)

r = 2, a = 5m + 2

a² = (5m + 2)² = 25m² + 4 + 20m = 5(5m² + 4m) + 4 = 4q + 4, where q = 5m² + 4m

r = 3, a = 5m + 3

a² = (5m + 3)² = 25m² + 9 + 30m = 5(5m² + 6m + 1) + 4

= 5q + 4, where q = 5m² + 6m + 1

r = 4, a = 5m + 4

a² = (5m + 4)² = 25m² + 16 + 40m = 5(5m² + 8m + 3) + 1

= 5q + 1, where q = 5m² + 8m + 3

Therefore, the square of any positive integer is of the form 5q, 5q + 1,5q + 4 for some integer q.

Question 10. Show that the square of odd integer is of the form 8q + 1, for some integer q.

Solution:

a = bq+r , 0 < r < b (by Euclid's lemma)

Putting b=4 for the question,

⇒ a = 4q + r, 0 < r < 4

For r = 0, a = 4q, which is an even number.

For r = 1, a = 4q + 1, which is an odd number.

On squaring,

⇒ a² = (4q + 1)² = 16q² + 1 + 8q = 8(2q² + q) + 1 = 8m + 1, where m = 2q² + q

For r = 2, a = 4q + 2 = 2(2q + 1), which is an even number.

For r = 3, a = 4q + 3, which is an odd number.

On squaring,

⇒ a² = (4q + 3)² = 16q² + 9 + 24q = 8(2q² + 3q + 1) + 1

= 8m + 1, where m = 2q² + 3q + 1

Therefore, the square of an odd integer is of the form 8q + 1, for some integer q.

Practice Questions on Real Numbers

1).Prove that √2 is an irrational number.

2).Express 0.1363636... as a rational number in its simplest form.

3).Find the HCF of 336 and 54 using the Euclidean division algorithm.

4).Show that any positive odd integer is of the form 4q+1 or 4q+3, where q is some integer.

5).Prove that √3 is an irrational number.

6).Express 2.3333... as a rational number in its simplest form.

7).If a = 3 + √5, find the value of a² - 2a - 4.

8).Prove that the square of any odd integer is of the form 8q + 1, where q is some integer.

9).Find the LCM of 120 and 144 using the prime factorization method.

10).If x = √2 + 1, express x² + 1/x² in terms of rational numbers.

Read more :

• Real-Life Applications of Real Numbers
• Real Numbers

Summary

Exercise 1.1 of Chapter 1 (Real Numbers) in RD Sharma's Class 10 mathematics textbook focuses on the fundamental concepts of real numbers, including rational and irrational numbers, their representations, and operations. This exercise helps students understand the properties of real numbers and how to work with them effectively.