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

Hence Proved!!

a and b are two odd positive integers then one number from $\frac{a+b}{2} , \frac{a-b}{2}$ is odd number and even number.

Step-by-step explanation:

Given:

a and b are two odd positive integers.

Now we know that;

The odd numbers are in the form of $2n+1$ and $2n+3$.

where n is a positive integer.

Now Given that;

$a>b$

So let us say that;

$a = 2n+3\\\\b= 2n+1$

We need to prove that one number from $\frac{a+b}{2} , \frac{a-b}{2}$ is odd number and even number.

Solution:

First we will find the $\frac{a+b}{2}$.

$\frac{a+b}{2} = \frac{2n+3+2n+1}{2}\\\\\frac{a+b}{2} = \frac{4n+4}{2}\\\\\frac{a+b}{2} = \frac{4(n+1)}{2}\\\\\frac{a+b}{2} = 2(n+1)$

Let us assume $n+1 = m$

$\frac{a+b}{2} = 2m$

Hence $\frac{a+b}{2}$ is an even Number.

Now we will find for $\frac{a-b}{2}$

$\frac{a-b}{2}= \frac{2n+3-(2n+1)}{2}\\\\\frac{a-b}{2}= \frac{2n+3-2n-1}{2}\\\\\frac{a-b}{2}= \frac{2}{2}\\\\\frac{a-b}{2}= 1$

Hence $\frac{a-b}{2}$ is an odd number.

Hence Proved!!!

Answer 2

Given:-

• If a and b are two odd positive integers such that a > b.

To prove:-

• That one of the two numbers (a + b)/2 and (a – b)/2 is odd and the other is even.

Proof:-

• Let a and b be any odd odd positive integer such that a > b

Since any positive integer is of the form q, 2q + 1

Let a = 2q + 1 and b = 2m + 1,

Where, q and m are some whole numbes.

=> a + b/2 = (2q + 1) - (2m + 1) / 2

=> a + b/2 = 2(q + m) + 1 / 2

=> a + b/2 = (q + m + 1) / 2

Which is a positive integer.

Also,

=> a - b/2 = (2q + 1) - (2m + 1) / 2

=> a - b/2 = 2(q - m) / 2

=> a - b/2 = (q - m)

Given a > b

Therefore,

=> 2q + 1 > 2m + 1

=> 2q > 2m

=> q = m

Therefore,

=> a + b / 2 = (q - m) > 0

Thus, (a + b)/2 is a positive integer.

Now,

Prove that one of the two numbers (a + b)/2 and (a – b)/2 is odd and the other is even.

=>(a + b)/2 - (a – b)/2

=> (a + b) - (a - b) / 2

=> 2b/2

=> b

b is odd positive integer.

Also,

The proof above that (a + b)/2 and (a - b)/2 are positive integers.

Hence, it is proved that if a and b are two odd positive integer such that a > b then one of the two numbers (a + b)/2 and (a - b)/2 is odd and the other is even.