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

Asked by  | 12th Jun, 2011, 09:39: PM

Expert Answer:

Let us suppose that the numbers (a+b)/2 and (a-b)/2 are either both odd or both even.

Case1) When (a+b)/2 and (a-b)/2 are odd. 

We know that the sum or difference of two odd numbers is even, hence the sum of the numbers (a+b)/2 and (a-b)/2 must be even. 

So, (a+b)/2 +(a-b)/2=a must be even which is not correct as we are given that a is odd positive integer.

(If we take the difference, we will get the value as equal to b).

This leads to a contradiction.  Hence (a+b)/2 +(a-b)/2 cannot be both odd.

Case 2) When (a+b)/2 and (a-b)/2 are even.

Again, the sum or difference of two even numbers is even. 

So, (a+b)/2 +(a-b)/2=a must be even, which is not correct as we are given that a is odd positive integer.  (If we take the difference, we will get value equal to b).  This again leads to a contradiction.  Hence (a+b)/2 +(a-b)/2 cannot be both even.

So, the given two numbers cannot be both even or both odd.  Hence, there is only one possibility that one out of a+b/2 and a-b/2 is odd and the other is even.

Answered by  | 12th Jun, 2011, 09:58: PM

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