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

**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)

= n

^{2 }+ nAny positive integer is of the form 2q or 2q+1.

Let n = 2q

⇒ n

^{2 }+ n = (2q)^{2 }+2q⇒n

^{2 }+ n = 4q^{2 }+2q⇒ n

^{2 }+ n = 2(2q^{2 }+q)Therefore, n

^{2 }+ n is divisible by 2.n = 2q+1

⇒ n

^{2 }+ n = (2q+1)^{2}+ (2q+1)⇒ n

^{2 }+ n = (4q^{2}+4q+1 +2q+1)⇒ n

^{2 }+ n = (4q^{2}+6q+2)⇒ n

^{2 }+ n = 2(2q^{2}+3q+1)Thus, n

^{2 }+ n is divisible by 2Therefore, 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 n**^{3} – n divisible by 6.

^{3}– 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 n

^{3}– n = n(n^{2}-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, n

^{3}– 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,

⇒ n

^{2}= (5q+1)^{2}⇒ n

^{2}= (25q^{2}+10q+1)⇒ n

^{2}= 5(5q^{2}+2q)+1⇒ n

^{2}= 5m+1, where m is some integer. [For m = 5q^{2}+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

⇒ n

^{2}= (3q)^{2 }= 9q^{2}⇒ n

^{2}= 3(3q^{2})⇒ n

^{2}= 3m, where m is some integer [m = 3q^{2}]If n= 3q+1,

Then, on squaring

⇒ n

^{2}= (3q+1)^{2 }= 9q^{2 }+ 6q + 1⇒ n

^{2}= 3(3q^{2 }+2q) + 1⇒ n

^{2}= 3m + 1, where m is some integer [m = 3q^{2 }+2q]If n= 3q+2,

Then, on squaring

⇒ n

^{2}= (3q+2)^{2 }= 9q^{2}+ 12q + 4⇒ n

^{2}= 3(3q^{2 }+ 4q + 1) + 1⇒ n

^{2}= 3m + 1, where m is some integer [m = 3q^{2 }+ 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

a

^{2}= 16p^{2}= 4(4p^{2}) = 4q, where q = 4p^{2}r = 1, a = 4p + 1

a

^{2}= (4p + 1)^{2}= 16p^{2}+ 1 + 8p = 4(4p + 2) + 1 = 4q + 1, where q = (4p + 2)r = 2, a = 4p + 2

a

^{2}= (4p + 2)^{2}= 16p^{2}+ 4 + 16p = 4(4p^{2}+ 4p + 1) = 4q, where q = 4p^{2}+ 4p + 1r = 3, a = 4k + 3

a

^{2}= (4p + 3)^{2}= 16p^{2}+ 9 + 24p = 4(4p^{2}+ 6p + 2) + 1= 4q + 1, where q = 4p

^{2}+ 6p + 2Therefore, 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

a

^{2}= 25m^{2}= 5(5m^{2}) = 5q, where q = 5m^{2}When r = 1, we get, a = 5m + 1

a

^{2}= (5m + 1)^{2}= 25m^{2}+ 1 + 10m = 5m(5m + 2) + 1 = 5q + 1, where q = m(5m + 2)r = 2, a = 5m + 2

a

^{2}= (5m + 2)^{2}= 25m^{2}+ 4 + 20m = 5(5m^{2}+ 4m) + 4 = 4q + 4, where q = 5m^{2}+ 4mr = 3, a = 5m + 3

a

^{2}= (5m + 3)^{2}= 25m^{2}+ 9 + 30m = 5(5m^{2}+ 6m + 1) + 4= 5q + 4, where q = 5m

^{2}+ 6m + 1r = 4, a = 5m + 4

a

^{2}= (5m + 4)^{2}= 25m^{2}+ 16 + 40m = 5(5m^{2}+ 8m + 3) + 1= 5q + 1, where q = 5m

^{2}+ 8m + 3Therefore, 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,

⇒ a

^{2}= (4q + 1)^{2}= 16q^{2}+ 1 + 8q = 8(2q^{2}+ q) + 1 = 8m + 1, where m = 2q^{2}+ qFor 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,

⇒ a

^{2}= (4q + 3)^{2}= 16q^{2}+ 9 + 24q = 8(2q^{2}+ 3q + 1) + 1= 8m + 1, where m = 2q

^{2}+ 3q + 1Therefore, the square of an odd integer is of the form 8q + 1, for some integer q.

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