If a and b are two odd positive integers such that a > b

Solution:

Given: If a and b are two odd positive integers such that a > b.

To Prove: That one of the two numbers $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and the other is even.

Proof: Let a and b be any odd odd positive integer such that a > b.

Since any positive integer is of the form q, 2q + 1

Let $a=2 q+1$ and $b=2 m+1$, where, $q$ and $m$ are some whole numbers

$\Rightarrow \frac{a+b}{2}=\frac{(2 q+1)+(2 m+1)}{2}$

$\Rightarrow \frac{a+b}{2}=\frac{2(q+m)+1)}{2}$

$\Rightarrow \frac{a+b}{2}=(q+m+1)$

which is a positive integer.
Also,

$\Rightarrow \frac{a-b}{2}=\frac{(2 q+1)-(2 m+1)}{2}$

$\Rightarrow \frac{a-b}{2}=\frac{2(q-m)}{2}$

$\Rightarrow \frac{a-b}{2}=(q-m)$

Given, a > b

$\therefore 2 q+1>2 m+1$

$\Rightarrow 2 q>2 \mathrm{~m}$

$\Rightarrow q>m$

$\therefore \frac{a-b}{2}=(q-m)>0$

Thus, $\frac{(a-b)}{2}$ is a positive integer.

Now, we need to prove that one of the two numbers $\frac{(a+b)}{2}$ and $\frac{(a-b)}{2}$ is odd and other is even.

Consider, $\frac{(a+b)}{2}-\frac{(a-b)}{2}=\frac{(a+b)-(a-b)}{2}=\frac{2 b}{2}=b$, which is odd positive integer.

Also, we know from the proof above that $\frac{(a+b)}{2}$ and $\frac{(a-b)}{2}$ are positive integers.

We know that the difference of two positive integers is an odd number if one of them is odd and another is even. (Also, difference between two odd and two even integers is even) 

Hence it is proved that If $a$ and $b$ are two odd positive integers such that $a>b$ then one of the two numbers $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and the other is even.