#### RD Sharma 10th class Chapter 1- Real Numbers Exercise 1.1 Page No: 1.10

**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:**

We know that any odd positive integer is of form 4q+1 or 4q+3 for some whole number q.

Now that it’s given a > b

So, we can choose a= 4q+3 and b= 4q+1.

∴ (a+b)/2 = [(4q+3) + (4q+1)]/2

⇒ (a+b)/2 = (8q+4)/2

⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number.

Now, doing (a-b)/2

⇒ (a-b)/2 = [(4q+3)-(4q+1)]/2

⇒ (a-b)/2 = (4q+3-4q-1)/2

⇒ (a-b)/2 = (2)/2

⇒ (a-b)/2 = 1, which is an odd number.

Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd, and the other is even**.**

**2. Prove that the product of two consecutive positive integers is divisible by 2.**

**RD Sharma 10th class Chapter 1- Real Numbers**

**Solution:**

Let’s consider two consecutive positive integers as (n-1) and n.

∴ Their product = (n-1) n

= n^{2 }– n

We know that any positive integer is of form 2q or 2q+1. (From Euclid’s division lemma for b= 2)

So, when n= 2q

We have,

⇒ n^{2 }– n = (2q)^{2 }– 2q

⇒ n^{2 }– n = 4q^{2 }-2q

⇒ n^{2 }– n = 2(2q^{2 }-q)

Thus, n^{2 }– n is divisible by 2.

Now, when n= 2q+1

We have,

⇒ n^{2 }– n = (2q+1)^{2} – (2q-1)

⇒ n^{2 }– n = (4q^{2}+4q+1 – 2q+1)

⇒ n^{2 }– n = (4q^{2}+2q+2)

⇒ n^{2 }– n = 2(2q^{2}+q+1)

Thus, n^{2 }– n is divisible by 2 again.

Hence, the product of two consecutive positive integers is divisible by 2.

**3.** **Prove that the product of three consecutive positive integers is divisible by 6.**

**RD Sharma 10th class Chapter 1- Real Numbers**

**Solution:**

Let n be any positive integer.

Thus, the three consecutive positive integers are n, n+1 and n+2.

We know that any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)

So,

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.

**4. For any positive integer n, prove that n ^{3} – n is divisible by 6.**

**Solution:**

Let n be any positive integer. And since any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 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)

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RD Sharma 10th class Chapter 1- Updated 2022-2023

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