Question

1. If a and bare two odd positive integers such that a > b, then prove that one of the
atb
numbers -
and
ab
- is odd and the other is even.


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Answer 1

Answer:

First we can easily verify that a+b/2 and a-b/2  are positive integers since the sum of two odd numbers is always even and, the difference of two odd numbers is always even respectively.

This implies that on division by  2 we will have a positive integer.

Let 

x=  a+b/2 + a-b/2

therefore x=a

Therefore, we have that  x is an odd positive integer. We know that the sum of two even or sum of two odd numbers is never odd. Thus, it follows that a+b/2 is even when a-b/2 is odd and vice-versa.

Hence proved.

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Answer 2

Answer:

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