Question

if a and b are two odd positive integer such that a>b then prove that one of the two number a+b÷2
and a-b÷2 is odd and the other is even



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Answer 1

Given ,

a and b are two odd positive integers .

We know that every odd integer is of the form 2m+1 . So , a and b are also of 2m+1 form .

Now let ,

a= 2x +1 and b= 2y + 1

where x and y are two positive integers

Therefore ,

a + b = 2x +1 +2y +1

=>a+b = 2x+2y+2

=>a + b = 2 (x+y+1)

=>(a+ b)÷2 = ( x +y+1)

Again given,

a>b

that mean x>y

a-b = 2x +1 -2y -1

=> a - b = 2x -2y

=> a - b = 2 ( x -y)

=> (a - b) ÷ 2= (x-y)

Since , x>y so (x- y ) is positive

When we take one as odd and other as even fron x and y we get

x - y is odd

and x +y +1 is even

In this case

(a+b)÷2 is even

while (a-b)÷2 is odd

again , when we take x and y both as rather odd or even we get

x - y is even

but x+y+1 is odd

In this case

(a+b)÷2 is odd

while (a-b) ÷2 is even

Hence it is proven that if a and b are two odd positive integers, then from (a+b)÷2 and (a-b)÷ 2 one is odd while other is even.

Here's an example to demonstrate :

We take two odd positive integers 5 and 7 .

(7+5 )÷ 2

= 12÷2

= 6 ( even )

And

(7- 5)÷ 2

=2÷2

=1 ( odd)

Answer 2

Answer.

Let a and b are any two odd positive integers.

Hence and where m and n are whole numbers.

Consider

Therefore is a positive integer.

Now,

But given a > b

Hence is also a positive integer

Now we have to prove that of the numbers and is odd and another is even number.

Consider,

which is an odd positive integer à (1)

It is already proved that and are positive integers à (2)

Recall that the difference between an odd number and even number is always an odd number.

Hence from (1) and (2), we can conclude that one of the integers and is even and other is odd.