If a and b are two odd positive integers such that a > b, prove that one of the numbers (a + b)/2 and (a − b)/2 is odd and the other is even.

In this problem, we prove that if a and b are two odd positive integers with
$a>b$, then the two numbers
$(a+b)/2$ and $(a-b)/2$ are such that one is odd and the other is even.
This type of proof is important in Number System and Mathematical Reasoning.

Question

If a and b are two odd positive integers such that a > b, prove that one of the two numbers (a + b)/2 and (a − b)/2 is odd and the other is even.

Solution

Since a and b are odd positive integers, they can be written in the form:

a = 2m + 1
b = 2n + 1

where m and n are integers.

Now,

(a + b)/2
= (2m + 1 + 2n + 1) divided by 2
= (2m + 2n + 2) divided by 2
= m + n + 1

Also,

(a − b)/2
= (2m + 1 − 2n − 1) divided by 2
= (2m − 2n) divided by 2
= m − n

Here, m + n + 1 and m − n are integers.

If m + n + 1 is odd, then m − n is even.
If m + n + 1 is even, then m − n is odd.

Therefore, one of the numbers (a + b)/2 and (a − b)/2 is odd and the other is even.

Conclusion

Hence, it is proved that if a and b are two odd positive integers such that a > b, then one of the numbers (a + b)/2 and (a − b)/2 is odd and the other is even.

Hence proved.

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