Question

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

Answer 1

If $a$ and $b$ are odd positive integers, they are one more than a non-negative even number, i.e. there exists $m,n \in \mathbb{N}$ such that

$a = 2m+1,\quad b = 2n+1$

So, the first expression become

$\cfrac{a+b}{2} = \cfrac{2m+1+2n+1}{2} = \cfrac{2m+2n+2}{2} = m+n+1$

Similarly, we have

$\cfrac{a-b}{2} = \cfrac{2m+1-2n-1}{2} = \cfrac{2m-2n}{2} = m-n$

Now, the parity of these expressions depend on those of $m$ and $n$. We have four cases:

If both m and n are even:

$m+n+1$ is odd, since $m+n$ is even, while $m-n$ is even

If one of the two is odd and the other is even:

$m+n+1$ is even, since $m+n$ is odd, while $m-n$ is odd

If both are odd:

$m+n+1$ is odd, since $m+n$ is even, while $m-n$ is even

So, in all cases, one between (a+b)/2 and (a-b)/2 is odd, and the other is even.