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Inequalities DS Question from OG

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Inequalities DS Question from OG

by arvn » Mon Sep 01, 2008 11:49 am
The question:

Is 1/(a-b) < (b-a) ?
(1) a < b
(2) 1 < |a-b|

I approached this problem by rephrasing the question as
1/(a-b) < -(a-b) ?
==> 1 < -(a-b)^2
which is never true as the square of any number is positive and when you multiply that by a negative sign the inequality become
IS 1 < -ve number? Which is never true. The question can be answered irrespective of what (1) and (2) says. I am little confused ..

What is the best approach? Can someone help me please?
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by bomond » Tue Sep 02, 2008 8:52 am
Hi arvn

You make some mistakes while calculations:

1/(a-b) < -(a-b) ?
==> 1 < -(a-b)^2

You can not above mentioned calculation while you are not sure a-b are positive. If it is positive you calculation is true. But if it is negative the ineq symbel changes itself. I mean it would be 1> -(a-b)^2

That's why it depends on a-b

Hope it helps
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by pkw209 » Thu May 06, 2010 4:16 pm
Is 1/(a-b) < (b-a) ?
Is there a reason why we can't rephrase this as "is 1 < (a-b)(b-a)?"
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by pkw209 » Thu May 06, 2010 4:19 pm
If the above rephrasing is correct / allowed, then wouldn't statement B always be no?
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