If $a$ and $b$ are two odd positive integers such that $a > b,$ then prove that one of the two numbers $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and other is even.

Since $a$ and $b$ are odd positive integers,we can represent them as $a = 2m + 1$ and $b = 2n + 1$ for some non-negative integers $m$ and $n$ where $m > n$.
Now,consider the sum: $\frac{a+b}{2} = \frac{(2m+1) + (2n+1)}{2} = \frac{2m + 2n + 2}{2} = m + n + 1$.
Next,consider the difference: $\frac{a-b}{2} = \frac{(2m+1) - (2n+1)}{2} = \frac{2m - 2n}{2} = m - n$.
Let $S = m + n + 1$ and $D = m - n$.
Consider the difference between these two results: $S - D = (m + n + 1) - (m - n) = 2n + 1$.
Since $2n + 1$ is an odd number,the difference between $S$ and $D$ is odd.
If the difference between two integers is odd,one must be even and the other must be odd.
Therefore,one of $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and the other is even.