Yes, you just flip the inequality sign. What you did with (a - b) is absolutely correct.
But, you have to understand why we flip it.

Let's take this example:

3 > 2
If you multiply both sides with a positive number, say 2,
6 > 4
So, the inequality sign does not change.

But, if you multiply both sides with a negative number say -2,
we can't write
- 6 > - 4
The sign should flip and
- 6 < - 4 is correct.

So remember, (as you did) when you multiply both sides of an inequality with a positive number the sign does not flip. But when you multiply both sides by a negative number, the sign of inequality flips.

The case of |x-1| > 2 is concerned with understanding modulus and opening the modulus.
|x - 1| opens up as (x - 1) when (x - 1) > 0
so (x - 1) > 2 when x > 1
|x - 1| opens up as -(x - 1) when (x - 1) < 0
so -(x - 1) > 2 or (x - 1) < -2 when x < 1

massi2884 wrote:Is 1/(a - b) < b - a ?

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

Statement 1: a < b
Thus, a-b<0, implying that b-a>0.
Is 1/(negative) < positive?
YES.
SUFFICIENT.

Statement 2: 1 < |a - b|
It's possible that a-b = 2, implying that b-a = -2.
Plugging a-b=2 and b-a=-2 into 1/(a-b) < b-a, we get:
1/2 < -2?
NO.

It's possible that a-b = -2, implying that b-a = 2.
Plugging a-b=-2 and b-a=2 into 1/(a-b) < b-a, we get:
1/-2 < 2?
YES.

Since in the first case the answer is NO but in the second case the answer is YES, INSUFFICIENT.

The correct answer is A.

Hi Mitch,
I found the insufficiency of 2 the following way. hope this is right

1< |a-b|

take x=a-b

=> x>1 or x<-1
since x cant be a distinct no, it is insufficient
Just hope the above reasoning is right..
Please correct me if I am wrong.

Hello experts,

Can we write the original statement 1/(a-b)<b-a as (a-b)^2<-1?

Regards,
Amit