Class 10 RD Sharma Solutions - Chapter 1 Real Numbers - Exercise 1.1 | Set 2

Chapter 1 of RD Sharma's Class 10 textbook focuses on the Real Numbers laying a foundational understanding of the number systems. Exercise 1.1 | Set 2 deals with the problems aimed at enhancing students' grasp of real numbers including their properties and classifications. This exercise is designed to build proficiency in handling the various types of real numbers and solving related problems.

Real Numbers

The Real numbers include all the numbers that can be found on the number line. This encompasses rational numbers and irrational numbers. Real numbers can be further categorized into different types including natural numbers, whole numbers, integers, and fractions. They are essential for performing the arithmetic operations and solving equations in the algebra and beyond.

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:

a = bq+r ; where 0 < r < b

Putting b=6 we get,

⇒ a = 6q + r, 0 < r < 6

r = 0, a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]

r = 1, a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]

r = 2, a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]

r = 3, a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]

r = 4, a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]

r = 5, a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]

Therefore, any odd positive integer can be of the form 6q +1,6q + 3,6q + 5, where q is some integer.

Question 12. Show that the square of any positive integer cannot be of form 6m + 2 or 6m + 5 for any integer m.

Solution:

a = 6q + r, where 0 ≤ r < 6 (Taking b=6 in Euclid's division lemma)

a² = (6q + r)² = 36q² + r² + 12qr 
a² = 6(6q² + 2qr) + r²  0 ≤ r < 6 

r = 0

a² = 6 (6q²) = 6m, where, m = 6q² is an integer.

r = 1

a² = 6 (6q² + 2q) + 1 = 6m + 1, where, m = (6q² + 2q) is an integer.

r = 2, 

a² = 6(6q² + 4q) + 4 = 6m + 4, where, m = (6q² + 4q) is an integer.

r = 3, 

a² = 6(6q² + 6q) + 9 = 6(6q² + 6q) + 6 + 3

a² = 6(6q² + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

r = 4,  

a² = 6(6q² + 8q) + 16

= 6(6q² + 8q) + 12 + 4

⇒ a² = 6(6q² + 8q + 2) + 4 = 6m + 4, where, m = (6q² + 8q + 2) is integer.

r = 5, 

a² = 6 (6q² + 10q) + 25 = 6(6q² + 10q) + 24 + 1

a² = 6(6q² + 10q + 4) + 1 = 6m + 1, where, m = (6q² + 10q + 4) is integer.

Therefore, 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 of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Solution:

For 6q,

(6q)³ = 216 q³ = 6(36q)³ + 0 

= 6m + 0, (where m is an integer = (36q)³)

For 6q+1,

(6q+1)³ = 216q³ + 108q² + 18q + 1 

= 6(36q³ + 18q² + 3q) + 1 

= 6m + 1, (where m is an integer = 36q³ + 18q² + 3q)

For 6q+2,

(6q+2)³ = 216q³ + 216q² + 72q + 8 

= 6(36q³ + 36q² + 12q + 1) +2 

= 6m + 2, (where m is an integer = 36q³ + 36q² + 12q + 1)

For 6q+3,

(6q+3)³ = 216q³ + 324q² + 162q + 27 

= 6(36q³ + 54q² + 27q + 4) + 3 

= 6m + 3, (where m is an integer = 36q³ + 54q² + 27q + 4)

For 6q+4,

(6q+4)³ = 216q³ + 432q² + 288q + 64 

= 6(36q³ + 72q² + 48q + 10) + 4 

= 6m + 4, (where m is an integer = 36q³ + 72q² + 48q + 10)

For 6q+5,

(6q+5)³ = 216q³ + 540q² + 450q + 125 

= 6(36q³ + 90q² + 75q + 20) + 5 

= 6m + 5, (where m is an integer = 36q³ + 90q² + 75q + 20)

Therefore, 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 form 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.

Solution:

b=5

n = 5q+r

0 < r < 5 

Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4

CASE 1:

When, n = 5q

n+4 = 5q+4

n+8 = 5q+8

n+12 = 5q+12

n+16 = 5q+16

n is only divisible by 5

CASE 2:

n = 5q+1

n+4 = 5q+5 = 5(q+1)

n+8 = 5q+9

n+12 = 5q+13

n+16 = 5q+17

n + 4 is only divisible by 5

CASE 3:

n = 5q+2

n+4 = 5q+6

n+8 = 5q+10 = 5(q+2)

n+12 = 5q+14

n+16 = 5q+18

n + 8 is only divisible by 5

CASE 4:

n = 5q+3

n+4 = 5q+7

n+8 = 5q+11

n+12 = 5q+15 = 5(q+3)

n+16 = 5q+19

n + 12 is only divisible by 5

CASE 5:

n = 5q+4

n+4 = 5q+8

n+8 = 5q+12

n+12 = 5q+16

n+16 = 5q+20 = 5(q+4)

Here, n + 16 is only divisible by 5

Therefore, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Question 15. Show that the square of an odd integer can be of the form 6q + 1 or 6q + 3, for some integer q.

Solution:

b=6

a = 6m + r 

0 ≤ r < 6.

a = 6m, 6m + 1, 6m + 2 , 6m + 3, 6m + 4, 6m + 5

Thus, we are choosing for a = 6m + 1 or, 6m + 3 or 6m + 5 for it to be an odd integer.

For a = 6m + 1,

(6m + 1)² = 36m² + 12m + 1

= 6(6m² + 2m) + 1

= 6q + 1, where q is some integer and q = 6m² + 2m.

For a = 6m + 3

(6m + 3)² = 36m² + 36m + 9

= 6(6m² + 6m + 1) + 3

= 6q + 3, where q is some integer and q = 6m² + 6m + 1

For a = 6m + 5,

(6m + 5)² = 36m² + 60m + 25

= 6(6m² + 10m + 4) + 1

= 6q + 1, where q is some integer and q = 6m² + 10m + 4.

Therefore, the square of an odd integer is of the form 6q + 1 or 6q + 3, for some integer q.

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.

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

⇒ a = 3q + r

So, a can be of the form 3q, 3q + 1 or 3q + 2.

Now, for a = 3q

(3q)² = 3(3q²) = 3m [where m = 3q²]

a = 3q + 1

(3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1 [where m = 3q² + 2q]

a = 3q + 2

(3q + 2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1

= 3m + 1 [where m = 3q² + 4q + 1]

Therefore, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.

Question 17. Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Solution:

a = bm + r

b = 3

a = 3m + r

r = 0, 1, 2.

r = 0, a = 3m.

r = 1, a = 3m + 1.

r = 2, a = 3m + 2.

When a = 3m

a² = (3m)² = 9m²

a² = 3(3m²) = 3q, where q = 3m²

When a = 3m + 1

a² = (3m + 1)² = 9m² + 6m + 1

a² = 3(3m² + 2m) + 1 = 3q + 1, where q = 3m² + 2m

When a = 3m + 2

a² = (3m + 2)²

a² = 9m² + 12m + 4

a² = 3(3m² + 4m + 1) + 1

a² = 3q + 1 where q = 3m² + 4m + 1

Therefore, square of any positive integer cannot be of the form 3q + 2, where q is a natural number.

Read More:

• Real Numbers
• Real-Life Applications of Real Numbers

Summary

Exercise 1.1 in Chapter 1 on Real Numbers for Class 10 typically delves into the fundamental concepts of the number system. It explores the various categories of numbers, including natural, whole, integer, rational, and irrational numbers, and how they collectively form the set of real numbers. This exercise usually focuses on the properties and representations of rational and irrational numbers, teaching students to distinguish between them and understand their decimal representations. Students learn to identify terminating and recurring decimals as rational numbers, while non-repeating, non-terminating decimals are recognized as irrational. The exercise often covers the representation of these numbers on a number line, emphasizing the density property of real numbers. It may also introduce operations on real numbers and their properties, helping students understand how different types of numbers interact under addition, subtraction, multiplication, and division. Through a series of problems and proofs, students are encouraged to deepen their understanding of the nature of real numbers, laying a crucial foundation for more advanced mathematical concepts they will encounter in higher studies.