hatemnag wrote:
Hi Bunuel
please help me in how to deal with the following expression: |x|−|y|<|x−y| in DS questions in simple way
Please read the entire thread to understand as it has been discussed at
is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html#p651477 and
is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html#p651854As for the theory or understanding behind the expression: |x|-|y| < |x-y|, you need to understand what do |x| and |x-y| mean in isolation.
|x| is the distance of 'x' from absolute 0 on the number line.
|x-y| represents the distance of 'x' from 'y' on the number line.
Thus, for S2, you are told that |x|-|y| < |x-y| ---> for such 'complex' absolute value inequality expressions, use numbers to guide you but for algebraic treatment, read along:
|x|-|y| < |x-y| ---> distance of x - distance of y < distance of x from y
Also, note that |any value| = always non-negative ---> |any number| \(\geq\) 0
Thus, |x| -|y| < a non negative quantity ---> |x| - |y| < 0 (I took 0 as it the simplest and the most straightforward example of a non-negative number).
2 cases on the number line that follow the fact that |x| - |y| < 0 :
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For case 1, you get xy<0 but for case 2, you get xy> 0 ---> S2 is NOT sufficient as it is giving us 2 different answers to the same question asked.
Hope this helps.