## 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

**Answer**: (C) 2m

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

**Answer**: (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

**Answer**: (C) an odd 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.

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