## Real Numbers (NCERT Exemplar Solutions Ex 1.1)

Question 1:
For some integer m, every even integer is of the form

(A) m
(B) m + 1
(C) 2m
(D) 2m + 1

Explanation:
Even integers are -4, -2, 0, 2, 4, 6 ...
Let m = integers[Since, here integer is represented by m]
such that m = ⋯ , −1, 0, 1, 2, 3, ...
∴ 2m = ⋯ , −2, 0, 2, 4, 6, ...
∴ even integers can be represented in the form of 2m

Question 2:
For some integer q, every odd integer is of the form
(A) q
(B) q + 1
(C) 2q
(D) 2q + 1

Explanation:

We know the odd integers are 1, 3, 5...

So, it can be written in the form of 2q + 1.
Where, q = integer

Or q = ⋯ , −1, 0, 1, 2, 3, ....
∴ 2q + 1  = ⋯ , −3, −1, 1, 3, 5, ...

Question 3:
n² − 1 is divisible by 8 if n is

(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer

Explanation:
Let a = n² − 1
Here n can be even or odd.

Case I:
n = Even i.e., n = 2k, where k is an integer.
⟹ a = (2k)² − 1
⟹ a = 4k² − 1

At k = −1,
a = 4(−1)² − 1 = 4 − 1 = 3, which is not divisible by 8.

At k = 0,
a = 4(0)² − 1 = 0 − 1 = −1, which is not divisible by 8.

Case II:
n = odd i.e., n = 2k + 1, where k is an integer
⟹ a = (2k + 1)² − 1
⟹ a = 4k² + 4k + 1 − 1
⟹ a = 4k² + 4k
⟹ a = 4k(k + 1)

At k = −1,
a = 4(−1)(−1 + 1) = 0 which is divisible by 8.
At k = 0,
a = 4(0)(0 + 1) = 0 which is divisible by 8.
At k = 1,
a = 4(1)(1 + 1) = 8 which is divisible by 8.

Hence, we can conclude from above two cases, if n is odd, then n² − 1 is divisible by 8.  