Given:

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

To prove:

Out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) one is odd and other is even.

Proof:

a and b both are odd positive integers,

Since,

\((\frac{a+b}{2})\) + \((\frac{a-b}{2})\) = \((\frac{a+b+a-b}{2})\) = a

Which is an odd number as "a" is given to be a odd number.

Hence one of the number out of \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) must be even and other must be odd because adding two even numbers gives an even number and adding two odd numbers gives an even number.

Here,

\(\frac{a+b}{2}\) will give an odd number and \(\frac{a-b}{2}\) will give an even number.

EXAMPLE:

Take a = 7 and b = 3 such that a > b

Now,

\(\frac{a+b}{2}\) = \(\frac{7+3}{2}\) = \(\frac{10}{2}\) = 5 which is odd

And \(\frac{a-b}{2}\) = \(\frac{7-3}{2}\) = \(\frac{4}{2}\) = 2 which is even

Conclusion:

Out of out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\), \(\frac{a+b}{2}\) is an odd number and \(\frac{a-b}{2}\) is an even number.