Question

1. If a and b are two odd positive integers such that a > b, then prove that one of the two
(a+b)/2 and (a-b)/2 is odd and other is even​

Answer 1

Answer:

Let, a = 2p+1, p∈N and b = 2q+1 , q∈N∪∪{0} ∵∵ a > b ⇒ 2p+1 > 2q+1 ⇒ p > q a+b2a+b2 = 2p+1+2q+122p+1+2q+12 = p+q+1 a−b2a−b2 = 2p+1−(2q+1)22p+1−(2q+1)2 = p-q Case-I: a+b2a+b2 is odd which implies p+q+1 is odd ⇒⇒ p+q is even (∵∵ odd - 1 = even) ⇒⇒ p+q-2q is even (∵∵ even - even = even) ⇒⇒ p-q is even ⇒⇒ a−b2a−b2 is even Case-II: a+b2a+b2 is even ⇒⇒ p+q+1 is even ⇒⇒ p+q is odd (∵∵ even-1 = odd) ⇒⇒ p+q-2q is odd (∵∵ odd - even = odd) ⇒⇒ p-q is odd ⇒⇒ a−b2a−b2 is odd.Read more on Sarthaks.com - https://www.sarthaks.com/26059/are-two-odd-positive-integers-such-that-then-prove-that-one-the-numbers-and-and-the-other-even

Step-by-step explanation:

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