arorag wrote:is 1/a-b < b-a ???

1. a<b
2.1> abs a-b

I think the above solutions are overcomplicating matters. From 1:

a < b, so b - a > 0 and a - b < 0.

If a-b is negative, so is 1/(a-b). So of course it's less than b-a, which is positive. Sufficient.

From 2 (which I've understood to say 1 > |a-b|), a-b could be positive or negative. Insufficient. A.

That said, if you did want to do this algebraically, from 1) you know that a-b<0. So if you multiply both sides of the inequality by (a-b), you need to reverse the inequality:

1/(a-b) < (b-a)
1 > (b-a)(a-b)

Then multiplying by -1, we again need to reverse the inequality:

-1 < (a-b)(a-b)
-1 < (a-b)^2

which must be true; (a-b)^2 can't be less than zero.