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.