Prove That One of (a+b)/2 and (a−b)/2 Is Odd and the Other Is Even

Video Explanation

Question

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

Solution

Step 1: Express Odd Integers in Standard Form

Since a and b are odd positive integers, we can write:

\[ a = 2m + 1,\quad b = 2n + 1 \]

where \(m\) and \(n\) are integers and \(m > n\).

Step 2: Find \(\frac{a+b}{2}\)

\[ \frac{a+b}{2} = \frac{(2m+1)+(2n+1)}{2} = \frac{2(m+n+1)}{2} = m+n+1 \]

Thus, \(\frac{a+b}{2}\) is an integer.

Step 3: Find \(\frac{a-b}{2}\)

\[ \frac{a-b}{2} = \frac{(2m+1)-(2n+1)}{2} = \frac{2(m-n)}{2} = m-n \]

Thus, \(\frac{a-b}{2}\) is also an integer.

Step 4: Determine Parity

The two numbers obtained are:

\[ m+n+1 \quad \text{and} \quad m-n \]

One of these is odd and the other is even because:

• If \(m+n\) is even, then \(m+n+1\) is odd and \(m-n\) is even.
• If \(m+n\) is odd, then \(m+n+1\) is even and \(m-n\) is odd.

Conclusion

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

\[ \therefore \quad \text{Proved.} \]