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 the other is even.

Given: a and b are two odd positive numbers such that a $>$ b.

To do: Here we have to prove that one of the two numbers $\frac{a\ +\ b}{2}$ and $\frac{a\ -\ b}{2}$ is odd and the other is even.

Solution:

We know that;

$a$ and $b$ are two odd positive integers such that $a\ >\ b$.

Also,

Odd numbers are in the form of $2n\ +\ 1$ and $2n\ +\ 3$, where $n$ is an integer.

As $a\ >\ b$, 

$a\ =\ 2n\ +\ 3$ and $b\ =\ 2n\ +\ 1$

Now,

Calculating value of  ($\frac{a\ +\ b}{2}$):

$\frac{a\ +\ b}{2}\ =\ \frac{2n\ +\ 3\ +\ 2n\ +\ 1}{2}$

$\frac{a\ +\ b}{2}\ =\ \frac{4n\ +\ 4}{2}$

$\frac{a\ +\ b}{2}\ =\ 2n\ +\ 2$

$\frac{a\ +\ b}{2}\ =\ 2(n\ +\ 1)$

Any number multiplied by 2 is even. So,  $2(n\ +\ 1)$  is even.

Therefore,

$\mathbf{\frac{a\ +\ b}{2}}$  is even.

Calculating value of  ($\frac{a\ -\ b}{2}$):

$\frac{a\ -\ b}{2}\ =\ \frac{2n\ +\ 3\ -\ 2n\ -\ 1}{2}$

$\frac{a\ -\ b}{2}\ =\ \frac{2}{2}$

$\frac{a\ -\ b}{2}\ =\ 1$

Which is odd.

Therefore,

$\mathbf{\frac{a\ -\ b}{2}}$  is odd.

Hence, we can see that one number is odd and other is even.