Greater than

[hey_thr67]

Is 1/(a-b)<(b-a)?

1. a<b
2. 1<|a-b|

OA is A


What is wrong with rephrasing the question to (b-a)^2 > 1

[Anurag@Gurome]

Is 1/(a-b)<(b-a)?

1. a < b
2. 1 < |a-b|

Statement 1: As a < b, (a - b) < 0
Hence, 1/(a - b) is negative but (b - a) is positive.
Hence, 1/(a - b) is always less than (b - a)

Sufficient

Statement 2: Consider the following two examples,
a = 3, b = 1 --> |a - b| > 1 ---> 1/(a - b) > (b - a) --> NO
a = 1, b = 3 --> |a - b| > 1 ---> 1/(a - b) < (b - a) --> YES

Not sufficient

The correct answer is A.

[Anurag@Gurome]

What is wrong with rephrasing the question to (b-a)^2 > 1

Because we cannot conclude that from the given condition.

Let us assume that 1/(a - b) is less than (b - a).

Now, for a < b,

• (a - b) < 0 and (b - a) > 0
  So, 1/(a - b) < (b - a)
  --> (b - a)(a - b) < 1 ............. Change of inequality sign as we are multiplying with a negative number
  --> -(a - b)(a - b) < 1
  --> -(a - b)² < 1
  
  Now, (a - b)² is a non-negative quantity. Hence, -(a - b)² is always less than or equal to zero. Therefore, -(a - b)² will be always less than 1.
  
  Hence, our initial assumption that 1/(a - b) < (b - a) is true for a < b.

And, for a > b,

• (a - b) > 0 and (b - a) < 0
  So, 1/(a - b) < (b - a)
  --> (b - a)(a - b) > 1 ............. No change of inequality sign as we are multiplying with a positive number
  --> -(a - b)(a - b) > 1
  --> -(a - b)² > 1
  
  Now, (a - b)² is a non-negative quantity. Hence, -(a - b)² is always less than or equal to zero. Therefore, -(a - b)² cannot be greater than 1.
  
  Hence, our initial assumption that 1/(a - b) < (b - a) must be false for a > b.