RD Chapter 1- Real Numbers Ex-1.2 |
RD Chapter 1- Real Numbers Ex-1.3 |
RD Chapter 1- Real Numbers Ex-1.4 |
RD Chapter 1- Real Numbers Ex-1.5 |
RD Chapter 1- Real Numbers Ex-1.6 |
RD Chapter 1- Real Numbers Ex-VSAQS |
RD Chapter 1- Real Numbers Ex-MCQS |

**Answer
1** :

a and b are two odd numbers such that a > b

Let a = 2n + 1, then b = 2n + 3

**Answer
2** :

Let n and n + 1 are two consecutive positive integer

We know that n is of the form n = 2q and n + 1 = 2q + 1

n (n + 1) = 2q (2q + 1) = 2 (2q2 + q)

Which is divisible by 2

If n = 2q + 1, then

n (n + 1) = (2q + 1) (2q + 2)

= (2q + 1) x 2(q + 1)

= 2(2q + 1)(q + 1)

Which is also divisible by 2

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

**Answer
3** :
Let n be the positive any integer Then

n(n + 1) (n + 2) = (n^{2} + n) (n + 2)

Which is also divisible by 6

Hence the product of three consecutive positive integers is divisible by 6

**Answer
4** :

Which is divisible by 6

Hence we can similarly, prove that n^{2} –n is divisible by 6 for any positive integer n.

Hence proved.

**Answer
5** :

Let n = 6q + 5, where q is a positive integer

We know that any positive integer is of the form 3k or 3k + 1 or 3k + 2, 1

q = 3k or 3k + 1 or 3k + 2

If q = 3k, then n = 6q + 5

**Answer
6** :

Let a be any positive integer

Then a = 5m + 1

a^{2} = (5m + 1 )^{2} = 25m^{2} +10m + 1

= 5 (5m^{2} + 2m) + 1

= 5q + 1 where q = 5m^{2} + 2m

Which is of the same form as given

Hence proved.

**Answer
7** :

Let a be any positive integer

Let it be in the form of 3m or 3m + 1

Let a = 3q, then

Hence proved.

**Answer
8** :

Let a be the positive integer and

Let a = 4m

Hence proved.

**Answer
9** :

Let a be the positive integer, and

Let a = 5m, then

**Answer
10** :

Let n is any positive odd integer

Let n = 4p + 1, then

(4p + 1)^{2} = 16p^{2} + 8p + 1

n^{2} = 8p (2p + 1) + 1

= 8q + 1 where q = p(2p + 1)

Hence proved.

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