Question

If a and bare two odd positive integers such that a > b, then prove that one of the two numbers $\frac{a+b}{2}$ and as $\frac{a-b}{2}$ is odd and the other is even.

Answer 1

Let $a$ and $b$ are any two odd $+ve$ integers

Hence, $a=2m+1$ and $b=2n+1$

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

$\therefore \frac{a+b}{2}$ is a positive integers.

Now, $\frac{a-b}{2}=\frac{(2m+1)-(2n+1)}{2}=\frac{2m-2n}{2}=(m-n)$

But, $a>b$

$\therefore (2m+1)>(2n+1)$

$\Rightarrow m>n$ or $m-n>0$

$\therefore \frac{a-b}{2}>0$

Hence, $\frac{a-b}{2}$ is also a positive integers.

Now, we have to prove that of the numbers $\frac{a+b}{2}$ and

$\frac{a-b}{2}$ is odd and another is even numbers.

Consider, $\frac{a+b}{2}-\frac{a-b}{2}$

$=\frac{a+b-a+b}{2}=\frac{2b}{2}=b$ which is odd positive integers .......(1)

It is already proved above that $\frac{a+b}{2}$ and $\frac{a-b}{2}$ are positive integer .......(ii)

But, the difference between odd number and even number is always an odd

number.

$\therefore$ From (i) and (ii) we can conclude that one of the integers

$\frac{a+b}{2}$ andee $\frac{a-b}{2}$ is even and other is odd.