If [tex] a [/tex] and [tex] b [/tex] are odd positive integers, they are one more than a non-negative even number, i.e. there exists [tex] m,n \in \mathbb{N} [/tex] such that
[tex] a = 2m+1,\quad b = 2n+1 [/tex]
So, the first expression become
[tex] \cfrac{a+b}{2} = \cfrac{2m+1+2n+1}{2} = \cfrac{2m+2n+2}{2} = m+n+1 [/tex]
Similarly, we have
[tex] \cfrac{a-b}{2} = \cfrac{2m+1-2n-1}{2} = \cfrac{2m-2n}{2} = m-n [/tex]
Now, the parity of these expressions depend on those of [tex] m [/tex] and [tex] n [/tex]. We have four cases:
If both m and n are even:
[tex] m+n+1 [/tex] is odd, since [tex] m+n [/tex] is even, while [tex] m-n [/tex] is even
If one of the two is odd and the other is even:
[tex] m+n+1 [/tex] is even, since [tex] m+n [/tex] is odd, while [tex] m-n [/tex] is odd
If both are odd:
[tex] m+n+1 [/tex] is odd, since [tex] m+n [/tex] is even, while [tex] m-n [/tex] is even
So, in all cases, one between (a+b)/2 and (a-b)/2 is odd, and the other is even.
