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
Wed May 10, 2017

i wanted to ask that rational numbers are in the form of p/q where q is not equal to zero. but we can write (root 2)/1. in this root 2 is p and 1 is q. q is not equal to zero. so<, it should be a rational number but always read in maths textbooks and vedio lessons of topperlearning that root 2 is irrational. why?

Sun April 23, 2017

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