Question

if a and b are two odd positive integers such that a>b , then prove that the one of the two numbers a+b/2 and a-b/ 2 is odd and the other is even​

Answer 1

Answer:

We have

a and b are two odd positive integers such that a & b

but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.

so, a=2n+3, b=2n+1, n∈1

Given ⇒ a>b

now, According to given question

Case I:

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

put let m=2n+1 then,

$\frac{a + b}{2} = 2m \: = > even \: number$

Case II:

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

$\frac{2}{2} = 1 \: = > odd \: number$

Hence we can see that, one is odd and other is even.

This is required solutions.

Answer 2

Answer:

We have

a and b are two odd positive integers such that a & b

but we know that odd numbers are in the form of 2n+1 and 2n+3 where n is integer.

so, a=2n+3, b=2n+1, n∈1

Given ⇒ a>b

now, According to given question

Case I:

\begin{gathered} \frac{a + b}{2} = \frac{2n + 3 + 2n + 1}{2} \\ = \frac{4n + 4}{2} \\ \: \ = 2n + 2 = 2(n + 1)\end{gathered}

2

a+b

=

2

2n+3+2n+1

=

2

4n+4

=2n+2=2(n+1)

put let m=2n+1 then,

\frac{a + b}{2} = 2m \: = > even \: number

2

a+b

=2m=>evennumber

Case II:

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

2

a−b

=

2

2n+3−2n−1

\frac{2}{2} = 1 \: = > odd \: number

2

2

=1=>oddnumber

Hence we can see that, one is odd and other is even.

This is required solutions.