Class 10 RD Sharma解决方案–第1章实数–练习1.1 |套装2

问题11。证明任何正奇数整数的形式为6q +1或6q + 3或6q + 5，其中q是某个整数。

解决方案：

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.

问题12。证明对于任何整数m，任何正整数的平方不能为6m + 2或6m + 5的形式。

解决方案：

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. 

问题13。证明正整数为6q + r，q为整数且r = 0、1、2、3、4、5的立方也为6m + r。

解决方案：

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.

问题14.证明n，n + 4，n + 8，n + 12和n + 16中只有一个可以被5整除，其中n是任何正整数。

解决方案：

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.

问题15：对于一个整数q，证明一个奇数整数的平方可以是6q +1或6q + 3的形式。

解决方案：

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.

问题16.正整数的形式为3q + 1，q是自然数。您是否可以将3m + 1、3m或3m + 2以外的任何形式的正方形写成某个整数m?证明你的答案。

解决方案：

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.

问题17.证明任何正整数的平方不能为3m + 2的形式，其中m是自然数。

解决方案：

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.