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CBSE Class 10 Answered

<div>If a and b are two positive odd integers prove that one out of a+b&divide;2 and a-b&divide;2 is odd and other is even</div>
Asked by shabnakader | 22 Jun, 2017, 06:24: PM
Expert Answer
Given that both a and b are positive odd integers.
Let a = 2j + 1 and b = 2k+ 1,  where j and k can be
i) both odd or
ii) both even or
iii) one odd and a > b
 
Now, if j and k are both odd and both even (cases i and ii) then,
begin mathsize 16px style fraction numerator straight a plus straight b over denominator 2 end fraction equals fraction numerator space 2 straight j plus 2 straight k plus 2 over denominator 2 end fraction space equals space straight j plus straight k plus 1 end style
Now j + k will be always even (sum of two odd number is even and sum of two even numbers is definitely even)
 
This means j + k + 1 is definitely odd begin mathsize 16px style rightwards double arrow fraction numerator left parenthesis straight a plus straight b right parenthesis over denominator 2 end fraction end style is definitely odd
 
and begin mathsize 16px style fraction numerator left parenthesis straight a minus straight b right parenthesis over denominator 2 end fraction equals fraction numerator space 2 straight j plus 1 minus left parenthesis 2 straight k plus 1 right parenthesis over denominator 2 end fraction equals space fraction numerator 2 left parenthesis straight j minus straight k right parenthesis over denominator 2 end fraction equals space straight j minus straight k end style
 
which will always be even since the difference of odd as well as even numbers is even.
 
Hence,  j - k is definitely evenbegin mathsize 16px style rightwards double arrow fraction numerator left parenthesis straight a minus straight b right parenthesis over denominator 2 end fraction end style is definitely even.
 
Now for case (iii) lets say j is odd and k is even
begin mathsize 16px style rightwards double arrow fraction numerator left parenthesis straight a plus straight b right parenthesis over denominator 2 end fraction equals space fraction numerator left parenthesis 2 straight j plus 2 straight k plus 2 right parenthesis over denominator 2 end fraction equals space straight j plus straight k plus 1. end style
Now j + k will be always odd (sum of one odd and one even number is always odd )
 
This means j + k + 1 is definitely even begin mathsize 16px style rightwards double arrow fraction numerator left parenthesis straight a plus straight b right parenthesis over denominator 2 end fraction end style is definitely even.
similar logic goes for begin mathsize 16px style fraction numerator left parenthesis straight a minus straight b right parenthesis over denominator 2 end fraction equals fraction numerator 2 left parenthesis straight j minus straight k right parenthesis over denominator 2 end fraction equals space straight j minus straight k end style which will definitely be odd.
Hence proved.
Answered by Rebecca | 22 Jun, 2017, 07:22: PM
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