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 If $a$ and $b$ are integers with $a > b$, what is the smallest possible positive value of \(\frac{a+b}{a-b} + \frac{a-b}{a+b}\)?

Guest Jul 29, 2017
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 #1
avatar+76929 
+1

 

(a + b)  / (a -b) +  (a - b) / (a + b)  →  [   (a + b)^2 +  ( a - b)^2 ]  / [ a^2 - b^2]  →

 

[ a ^ 2 + 2ab + b^2 + a^2 - 2ab + b^2) / [ a^2 - b^2]  →

 

[ 2a ^2 + 2 b^2]  / [ a^2  - b^2] →

 

2 [ a^2 + b^2] / [a^2 - b^2 ] → 

 

This expression will have the smallest possible value when the numerator of  a^2 + b^2  =  the denominator of a^2 - b^2

 

Note that when   a = any integer > 0 and  b  = 0   the numerator of  a^2 + b^2  =  the denominator of a^2 - b^2  = 1

 

So....the smallest positive value  is 2

 

 

cool cool cool

CPhill  Jul 29, 2017
edited by CPhill  Jul 29, 2017
edited by CPhill  Jul 29, 2017

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