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

Question

Is j-k > 1/j - 1/k ? 1. jk> -1 2. -1< k< j can I prove it using algebra and not picking number ?

yezz · Answer

plz revisit question is it 100j-k or 100(j-k) or the 100 is not there to start with

guerrero25 · Answer

plz revisit question is it 100j-k or 100(j-k) or the 100 is not there to start with

Edited . Sorry about the junk number .

yezz · Answer

is j-k > k-j / jk i.e. (j-k) - > 0 , is /jk > 0 , i.e is /jk > 0 , i.e. is [(J-K)(JK+1)\ JK > 0 from 1 -1

Zarrolou · Answer

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 Den>0 so the answer is YES pick -1

mau5 · Answer

Basically the question boils down to : Is (j-k)(1+jk)jk>0. From F.S 1, we have (1+jk)>0. Thus, we have to answer whether (j-k)jk>0. Also for jk>-1, we could have both j,k>0; both j,k<0 or j,k of the opposite signs.Insufficient. From F.S 2, we have (j-k)>0. Thus, we have to answer whether (1+jk)jk>0. This is possible only if jk>0 OR if jk<-1. Insufficient. Taking both together, we know that both (1+jk) and (j-k) are positive. Thus, we have to answer whether jk>0. Just as above, Insufficient. E.

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

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