guerrero25 wrote:

Is j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

can I prove it using algebra and not picking number ?

\(j-k > \frac{1}{j} - \frac{1}{k}\) equals \(\frac{jk(j-k)-(k-j)}{jk}>0\) don't divide by jk because it could be 0

this is true if N+ and D+ or N- and D-

1. jk> -1 D could be -ve or +ve, no info about N. Not sufficient

2. -1< k< j j-k>0 and k-j<0 Num= \(jk(+ve)-(-ve)\)

jk could be +ve or -ve , Not sufficient

1+2 The numerator is jk(j-k)-(k-j) and from 2 we know that j-k>0 and k-j<0 so Num= \(jk(+ve)-(-ve)=\) and the Den is jk> -1.

Pick jk>0: Num>0 [(+ve)(+ve)-(-ve)>0] Den>0 so the answer is YES

pick -1<jk<0: Den<0 and there are no douts here, but what about the N? is N=(-ve)(+ve)-(-ve) a negative number? it depends, we cannot say.

E

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