Question

is a and b are two odd positive integer 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.

Answer 1

Let us consider that (a+b)/2 and (a-b)/2 are both odd and both even

 

In the first case, let them be 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 value equal to b). This leads to a contradiction. Hence (a+b)/2 +(a-b)/2 cannot be both odd.

 

In the second case, let the numbers (a+b)/2 +(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 leads to a contradiction. Hence (a+b)/2 +(a-b)/2 cannot be both even.

 

So, the two numbers in question cannot be both even or both odd. Hence they have only one possibility left – one is even and the other is odd

Answer 2

$\huge \fbox \red{Solution:}$

We know that any odd positive integer is of the form 4q+1 or, 4q+3 for some whole number q.

Now that it’s given a > b

So, we can choose a= 4q+3 and b= 4q+1.

∴ (a+b)/2 = [(4q+3) + (4q+1)]/2

⇒ (a+b)/2 = (8q+4)/2

⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number.

Now, doing (a-b)/2

⇒ (a-b)/2 = [(4q+3)-(4q+1)]/2

⇒ (a-b)/2 = (4q+3-4q-1)/2

⇒ (a-b)/2 = (2)/2

⇒ (a-b)/2 = 1 which is an odd number.

Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even.