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.
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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.
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