We presure you are talking about (a+b)/2 and (a-b)/2
To start with both a and b are odd
Let a = 2j + 1
and b =2 k+ 1
where j and k can be
I) both odd
II) both even or
III) one odd
if j and k are both odd and both even (case I and II) then
(a+b)/2 = (2j+2k+2)/2 = j+k+1. Now j+k will be always even (sum of two odd number is even and sum of two even numbers is definitely even)
This means j+k+1 is defintely odd => (a+b)/2 is definitely odd
and (a-b)/2= (2j+1-(2k+1))/2= 2(j-k)/2 = j-k. Which will always be even as difference of odd as well as even numbers is even.
hence j-k is defintely even => (a-b)/2 is definitely even.
Now for case III lets say j is odd and k is even
(a+b)/2 = (2j+2k+2)/2 = j+k+1. Now j+k will be always odd (sum of one odd and one even number is always odd )
This means j+k+1 is defintely even => (a+b)/2 is definitely even
similar logic goes for (a-b)/2 = 2(j-k)/2 = j-k which will definitely be odd.