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 or a-b/2 is odd and other is even.

Answer 1

Answer:

they are proved

Step-by-step explanation:

if a=4x+1 (odd number will be in the form = x+1)

b=2x+1

a+b/2={(4x+1)+(2x+1)}/2

=6x+2/2

=3x+1

a-b/2={(4x+1)-(2x+1)}/2

=2x/2

=x

if x =even number a+b/2 will be odd

if x =odd number a+b/2 will be 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.