If a and b are two odd positive integers such that a > b, then prove t

1.	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.
Answer» 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. ∴ \(\frac{(a+b)}{2}\) = \(\frac{[(4q+3) + (4q+1)]}{2}\) ⇒ \(\frac{(a+b)}{2}\) = \(\frac{(8q+4)}{2}\) ⇒ \(\frac{(a+b)}{2}\) = 4q+2 = 2(2q+1) which is clearly an even number. Now, doing \(\frac{(a-b)}{2}\) ⇒ \(\frac{(a-b)}{2}\) = \(\frac{[(4q+3)-(4q+1)]}{2}\) ⇒ \(\frac{(a-b)}{2}\) = \(\frac{(4q+3-4q-1)}{2}\) ⇒ \(\frac{(a-b)}{2}\) = \(\frac{(2)}{2}\) ⇒ \(\frac{(a-b)}{2}\) = 1 which is an odd number. Hence, one of the two numbers \(\frac{(a+b)}{2}\) and \(\frac{(a-b)}{2}\) is odd and the other is even.

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.

Answer»

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.

∴ \(\frac{(a+b)}{2}\) = \(\frac{[(4q+3) + (4q+1)]}{2}\)

⇒ \(\frac{(a+b)}{2}\) = \(\frac{(8q+4)}{2}\)

⇒ \(\frac{(a+b)}{2}\) = 4q+2 = 2(2q+1) which is clearly an even number.

Now, doing \(\frac{(a-b)}{2}\)

⇒ \(\frac{(a-b)}{2}\) = \(\frac{[(4q+3)-(4q+1)]}{2}\)

⇒ \(\frac{(a-b)}{2}\) = \(\frac{(4q+3-4q-1)}{2}\)

⇒ \(\frac{(a-b)}{2}\) = \(\frac{(2)}{2}\)

⇒ \(\frac{(a-b)}{2}\) = 1 which is an odd number.

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