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can I prove it using algebra and not picking number ?

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

from 1

-1<jk<0 JK not = 0

i.e jk is -ve fraction, now to evaluate the question we ve ( jk+1) +ve and jk -ve , and thus we need to know j, k individually to evaluate (j-k).... no enough info ... insuff

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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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.