What are the values of n that satisfy the condition 1/|n|>n

IMO Answer is A

Explanation :
Mod (n) is always positive i.e. modulus of any integer is always positive.
Now the question asks to find conditions where 1/ Mod (N) > N

1 ) Consider case when 0<N<1,
Here N is positive proper fraction,

Assume N = 1/2, then
1/ Mod (1/2) = 1/ (1/2) = 2
2>1;
hence 1/ Mod (N) > N

2) Consider case when -infinity < N < 0

Assume N = -1, then Assume N = -1/2, then
1/ Mod (-1) = 1/1 = 1 1/Mod (-1/2) = 1/ (1/2) = 2
1 > -1; 2 > -1/2
hence 1/ Mod (N) > N hence 1/ Mod (N) > N


In short whenever a question does not state explicitly that a number is either positive or negative and integer or fraction, test the following numbers -2, -3/2, -1, -1/2, 0, 1/2, 1, 2, 3/2. Ensure you apply the restrictions given in the problem.

Use "AND" when all the test numbers in the given restrictions are true. Use "OR" when either case is true. Here we are sure that all numbers in Option A satisfy the given condition hence the "and" option i.e. Option A is correct.

Hi All,

While the general concepts involved in this question WILL show up on your GMAT, this prompt is poorly-written (and Answers D and E are the SAME answer, just written differently). There are plenty of properly-written questions that you can practice inequalities, ranges, absolute value, etc. on, so I would suggest that you skip this question.

GMAT assassins aren't born, they're made,
Rich

Hello

Can someone please clarify this sum, its still not clear to me.

Thanks.

You are asked about the values that satisfy \(\frac{1}{|n|}>n\)

The steps are as follows:

Realize that |a|=a for a\(\geq\) 0 and
|a|=-a for a<0.

In the given question, it is implied that n \(\neq\) 0 as this will make 1/|n| undefined.

Thus, \(\frac{1}{|n|}>n\) ---> consider 2 cases:

1. For n>0 : |n|=n --> \(\frac{1}{|n|}>n\) --> \(\frac{1}{n}>n\) ---> n^2<1 (you could multiply through by n as n>0) ---> n^2-1<0 --> -1<n<1 but as n>0, this gives you the common range of n as 0<n<1.

2. for n<0 : |n|=-n ---> \(\frac{1}{|n|}>n\) -->\(\frac{1}{-n}>n\) ---> \(\frac{1}{-n}-n>0\) (do not multiply throughout by n as it is now a negative number (n<0) and as such the sign of the inequality changes when you multiply by a negative number.

Thus you get, \(\frac{1}{-n}-n>0\) --> \(\frac{-1-n^2}{n}>0\) ---> \(\frac{1+n^2}{n}<0\), as n^2+1 is always >0 , \(\frac{1+n^2}{n}<0\) gives n<0.

Thus the common range for this scenario is n<0.

Combining both sets of ranges (from 1 and 2) , you get n<0 and 0<n<1.

Hence A is the correct answer.

Hope this helps.

Thanks Engr2012 for the reply. Understood both the cases.
However I am having trouble understanding these inequalities especially with mods.
In case if I multiply with -1 then inequality should be 1<-n^2 -- Why cant I solve this way?

I understand the ranges taken for quadratic equations & e.gs from mathworkbook but this problem with reciprocal & mod seems to be different.

Others in the group- please also pitch in for help.

It is because you can not multiply an inequality by a negative number without reversing the inequality. This is a very important rule to remember.

If you are given that a<b and you multiple both sides by -1, then it becomes -a>-b and NOT -a<-b. This is because if lets say a=4 and b=5, then you have a<b but -a=-4 and -b=-5 , in this case you end up getting -4 > -5.

This is the reason why you can not solve this question (or any inequality question) by multiplying with -1 and NOT changing the sign of inequality.

Additonally, if lets say what you are mentioning is correct, then you get \(-n^2>\)1 ---> \(n^2<\)-1 NOT POSSIBLE as \(n^2\) will always be \(\geq\) 0

Hope this helps.

Thanks Engr2012 for the detailed explanation, i went thru inequality basics again , got the concept.

When we get value n<0 is it equivalent to -infinity <n< 0?

[quote="seemachandran"]Thanks Engr2012 for the detailed explanation, i went thru inequality basics again , got the concept.

When we get value n0 is 0>n>+inf

Hope this helps.

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