Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 2) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 1.11

Q.11. Show that any positive odd integer is of the form 6q + 1 or, 6q + 3 or, 6q + 5, where q is some integer.
Ans. To Show: That any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is any some integer.
Proof: Let a be any odd positive integer and b = 6.
Then, there exists integers q and r such that
a = 6q + r, 0 ≤ r < 6 (by division algorithm)
⇒ a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4
But 6q or 6q + 2 or 6q + 4 are even positive integers.
So, a = 6q + 1 or 6q + 3 or 6q + 5
Hence it is proved that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is any some integer. 

Q.12. Show that the square of  any positive integer cannot be of the form 6m + 2  or 6m + 5  for any integer m.
Ans. Suppose a be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exists non-negative integers a and r such that
a = 6q + r, where 0≤r<6 
⇒ a² = (6q + r)² = 36q² + r² + 12qr
⇒ a² = 6(6q² + 2qr) + r² ...(1) where, 0 ≤ r < 6
Case: 1
When r = 0.
Putting r = 0 in (1), we get 
a² = 6(6q)² = 6m
Where, m = 6q² is an integer
Case: 2
When r = 1.
Putting r = 1 in (1), we get
a2 = 6(6q² + 2q) + 1
⇒ a² = 6m + 1
where, m = (6q² + 2q) is an integer
Case: 3
When r = 2.
Putting r = 2 in (1), we get 
a² = 6(6q² + 4q) + 4
⇒ a² = 6m + 4
where, m = (6q² + 4q) is an integer
Case: 4
When r = 3.
Putting r = 3 in (1), we get
a² = 6(6q² + 6q) + 9
⇒ a² = 6(6q² + 6q) + 6 + 3
⇒ a² = 6(6q² + 6q + 1) + 3
⇒ a² = 6m + 3
where, m = (6q² + 6q + 1) is an integer
Case: 5
When r = 4
Putting r = 4 in (1), we get
a² = 6(6q² + 8q) + 16
⇒ a² = 6(6q² + 8q) + 12 + 4
⇒ a² = 6(6q² + 8q + 2) + 4
⇒ a² = 6m + 4
where, m = (6q² + 8q + 2) is an integer
Case: 6
When r = 5.
Putting r = 5 in (1), we get
a² = 6(6q² + 10q) + 25
⇒ a² = 6(6q² + 10q) + 24 + 1
⇒ a² = 6(6q² + 10q + 4) + 1
⇒ a² = 6m + 1
where, m = (6q² + 10q + 1) is an integer
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5.

Q.13. Show that the cube of a positive  integer is of the form 6q + r, where q is an a integer and r = 0, 1, 2, 3, 4, 5.
Ans. Suppose a be any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers a and 6, there exists non-negative integers a and r such that
a = 6q + r, where 0 ≤ r < 6
⇒ a³ = (6q + r)³ = 216q³ + r³ + 3 x 6q x r(6q + r)⇒ a³ = 6(216q³ + 108q²r + 18qr²) + r³ ....(1) where, 0 ≤ r < 6
Case: 1
When r = 0.
Putting r = 0 in (1), we get
a³ = 216q³ = 6(36q³) = 6m
where, m = 36q³ is an integer
Case: 2
When r = 1.
Putting r = 1 in (1), we  get 
a³ = (216q³ + 108q² + 18q) + 1
⇒ a³ = 6(36q³ + 18q² + 3q) + 1
⇒ a³ = 6m + 1
where, m = 36q³ + 18q² + 3q) is an integer
Case: 3
When r = 2.
Putting r = 2 in (1), we get 
a³ = (216q³ + 216q² + 72q) + 8
a³ = (216q³ + 216q² + 72q + 6) + 2
⇒ a³ = 6(36q³ + 36q² + 12q + 1) + 2
⇒ a³ = 6m + 2
where, m = (36q³ + 36q² + 12q + 1) is an integer
Case: 4
When r = 3.
Putting r = 3 in (1), we get 
a³ = (216q³ + 324q² + 162q) + 27
a³ = (216q³ + 324q² + 162q + 24) + 3
⇒ a³ = 6(36q³ + 54q² + 27q + 4) + 3
⇒ a³ = 6m + 3
where, m = (36q³ + 54q² + 27q + 4) is an integer
Case: 5
When r = 4
Put r = 4 in (1), we get
a³ = (216q³ + 432q² + 288q) + 64
a³ = (36q³ + 72q² + 48q + 60) + 4
⇒ a³ = 6(36q³ + 72q² + 48q + 10) + 4
⇒ a³ = 6m + 4
where, m = (36q³ + 72q² + 48q + 10) is an integer
Case: 6
When r = 5.
Putting r = 5 in (1), we get
a³ = 216q³ + 540q² + 450q + 125
a³ = 6(36q³ + 90q² + 75q + 20) + 5
⇒ a³ = 6m + 5
where, m = (36q³ + 90q² + 75q + 20) is an integer
Hence, the cube of any positive integer of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Q.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.
Ans. Consider the numbers n, (n + 4), (n + 8), (n + 12) and (n + 16),
Suppose n = 5q + r, where 0 ≤ r < 5
n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4
(By Euclid's division algorithm)
Case: 1 
When n = 5q.
n = 5q is divisible by 5.
n + 4 = 5q + 4 is not divisible by 5.
n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3 is not divisible by 5.
n + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5.
n + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5.
Case: 2 
When n = 5q + 1.
n = 5q + 1 is not divisible by 5.
n + 4 = 5q + 1 + 4 = 5(q + 1) is divisible by 5.
n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4 is not divisible by 5.
n + 12 = 5q + 1 + 12 = 5(q + 2) + 3 is not divisible by 5.
n + 16 = 5q + 1 + 16 = 5(q + 3) + 2 is not divisible by 5.
Case: 3 
When n = 5q + 2.
n = 5q + 2 is not divisible by 5.
n + 4 = 5q + 2 + 4 = 5(q + 1) + 1 is not divisible by 5.
n + 8 = 5q + 2 + 8 = 5(q + 2) is divisible by 5.
n + 12 = 5q + 2 + 12 = 5(q + 2) + 4 is not divisible by 5.
n + 16 = 5q + 2 + 16 = 5(q + 3) + 3 is not divisible by 5.
Case: 4 
When n = 5q + 3.
n = 5q + 3 is not divisible by 5.
n + 4 = 5q + 3 + 4 = 5(q + 1) + 2  is not divisible by 5.
n + 8 = 5q + 3 + 8 = 5(q + 2) + 1 is not divisible by 5.
n + 12 = 5q + 3 + 12 = 5(q + 3) is divisible by 5.
n + 16 = 5q + 3 + 16 = 5(q + 3) + 4 is not divisible by 5.
Case: 5 
When n = 5q + 4.
n = 5q + 4 is not divisible by 5.
n + 4 = 5q + 4 + 4 = 5(q + 1) + 3  is not divisible by 5.
n + 8 = 5q + 4 + 8 = 5(q + 2) + 2 is not divisible by 5.
n + 12 = 5q + 4 + 12 = 5(q + 3) + 1 is not divisible by 5.
n + 16 = 5q + 4 + 16 = 5(q + 4) is divisible by 5.
Hence, in each case, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Q.15. Show that the square of an odd positive integer can be of the form 6q + 1 or  6q + 3 for some integer q.
Ans. It is known that any positive integer can be written in the form of 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some integer m.
Thus, an odd positive integer can be of the form  6m + 1,6m + 3,6m + 5.
We have, (6m + 1)² = 36m² + 12m + 1 = 6(6m² + 2m) + 1 = 6q + 1, where q = 6m² + 2m is an integer
(6m + 3)² = 36m² + 36m + 9 = 6(6m² + 6m + 1) + 3 = 6q + 3, where q = 6m² + 6m + 1 is an integer
(6m + 5)² = 36m² + 60m + 25 = 6(6m² + 10m + 4) + 1 = 6q + 1, where q = 6m² + 10m + 4 is an integer
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.

Q.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.
Ans. By Euclid's lemma, b = aq + r, 0 ≤ r < a.
Here, b is a positive integer and a = 3.
∴ b = 3q + r, for 0 ≤ r < 3
This must be in the form 3q, 3q + 1 or 3q + 2.
Now,
(3q)² = 9q² = 3m, where m=3q²
(3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + m+1, where m = 3q² + 4q + 1
Therefore, the square of a positive integer 3q + 1 is always in the form of 3m or 3m + 1 for some integer m.

Q.17. Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.
Ans. By Euclid's lemma, b = aq + r, 0 ≤ r < a.
Here, b is a positive integer and a = 3.
∴ b = 3q + r, for 0 ≤ r < 3
This must be in the form 3q, 3q + 1 or 3q + 2.
Now,
(3q)² = 9q² = 3m, where m = 3q²
(3q + 1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q
(3q + 2)² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1
Therefore, the square of a positive integer cannot be of the form 3m + 2, where m is a natural number.