Question

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

Answer 1

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 is even (\(\because\) even - even = even)

\(\Rightarrow\) p-q is even

\(\Rightarrow\) \(\frac{a-b}{2}\) is even

Case-II: \(\frac{a+b}{2}\) is even

\(\Rightarrow\) p+q+1 is even

\(\Rightarrow\) p+q is odd (\(\because\) even-1 = odd)

\(\Rightarrow\) p+q-2q is odd (\(\because\) odd - even = odd)

\(\Rightarrow\) p-q is odd

\(\Rightarrow\) \(\frac{a-b}{2}\) is odd.

Answer 2

Let a = 2q + 3 and b = 2q + 1 be two positive odd integers such that a > b

Hence one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even for any two positive odd integer​