n^2 - 1 is divisible by 8, if n is (A) an integer (B) a natural number (C) an odd integer (D) an even integer

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}  B ={x: x is an even natural number} = {2, 4, 6, 8 …}  C = {x:...

Y are you taking some odd integers as 4m+1 or 4m+3. Why no we take 2m+1 or 3m+1

Solution: Since x and y are odd positive integers, we have x = 2m + 1 and y = 2n + 1 => x2 + y2 = (2m + 1)2 + (2n +...

Let, a = 2p+1, p∈N and b = 2q+1 , q∈N\(\cup\){0} \(\because\) a > b ⇒ 2p+1 > 2q+1 ⇒ p > q \(\frac{a+b}{2}\) = \(\frac{2p+1+2q+1}{2}\) = p+q+1 \(\frac{a-b}{2}\) = \(\frac{2p+1-(2q+1)}{2}\) = p-q Case-I: \(\frac{a+b}{2}\) is odd which implies p+q+1 is odd \(\Rightarrow\) p+q is even (\(\because\) odd - 1 = even) \(\Rightarrow\) p+q-2q...

Because every odd integer will cover by 4m+1 or 4m+3. You may also use 2m+1.

(i) A set of odd natural numbers divisible by 2 is a null set because no odd number is divisible by 2. (ii) A set of even prime numbers is not...

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …} B ={x: x is an even natural number} = {2, 4, 6, 8 …} C = {x:...