Question
Sun June 12, 2011 By:

# 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.

Sun June 12, 2011

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.

Related Questions
Sun April 23, 2017

# Q1 - A positive integer is of the form 3q+1 , q being a natural number . Can you write its square in any form other than 3m+1 , 3m or 3m +2 for some integer m ? Justify your answer.      Q2 - The solution of RdSharma book - level 2 is not provided . Plz provide the answers .

Sat April 22, 2017