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

GivEn:-

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

To pr0vE:-

• One of the two number $\sf \dfrac{a - b}{2}$ and $\sf \dfrac{a + b}{2}$ is odd and the other is even.

Soluti0n:-

✦ 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

a > b ____[∵ GivEn]

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According to question:-

• Case I :-

$:\implies\sf \dfrac{a + b}{2} \\\\ :\implies\sf \dfrac{2n + 3 + 2n + 1}{2} \\\\ :\implies\sf \dfrac{4n + 4}{2} \\\\ :\implies\sf 2n + 2 \\\\ :\implies\sf 2(n + 1) \\\\ \sf Let\;m = 2(n + 1)\;then, \\\\ :\implies\sf \dfrac{a + b}{2} = 2m = \bf{(Even\;Number.)}$

• Case II :-

$:\implies\sf \dfrac{a - b}{2} \\\\ :\implies\sf \dfrac{ \cancel{2n} + 3 \cancel{- 2n} - 1}{2} \\\\ :\implies\sf \dfrac{2}{2} = \bf{(Odd\;Number.)}$

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

${\underline{\underline{\sf{\red{\dag\;Hence\;ProvEd!!}}}}}$

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