Question

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 1

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.

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.