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

问题1.如果a和b是两个奇数正整数，使得a> b，则证明两个数(a + b)/ 2和(a – b)/ 2中的一个是奇数，而另一个是偶数。

解决方案：

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.

问题2。证明两个连续正整数的乘积可被2整除。

解决方案：

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

问题3.证明三个连续正整数的乘积可被6整除。

解决方案：

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.

问题4.对于任何正整数n，请证明n ³ – n可被6整除。

解决方案：

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.

问题5.证明如果正整数的形式为6q + 5，那么对于某个整数q，形式为3q + 2，但反之则不。

解决方案：

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.

问题6.证明任何形式为5q +1的正整数的平方都具有相同的形式。

解决方案：

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.

问题7.证明任何正整数的平方形式为3m或3m +1，但形式不为3m + 2。

解决方案：

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.

问题8.证明对于某个整数q，任何正整数的平方均采用4q或4q +1的形式。

解决方案：

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.

问题9.证明某个正整数q的正整数的平方的形式为5q或5q + 1、5q + 4。

解决方案：

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.

问题10。表明奇数整数的平方形式为8q +1，对于某个整数q。

解决方案：

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.