Q1. If a and b are two odd positive integers such that a > b, then prove that one of the two numbers \(\frac{a+b}{2} \) and \(\frac{a-b}{2} \) is odd and the other is even.
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.
∴ \( \frac{a+b}{2}=\frac{\left[ (4q+3)+(4q+1) \right]}{2} \)
⇒ \(\Rightarrow \frac{a+b}{2}=\frac{8q+4}{2} \)(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
⇒ \(\Rightarrow \frac{a-b}{2}=\frac{\left[ (4q+3)-(4q+1) \right]}{2} \)
⇒ (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.
2
