If N is a negative, which of the following must be true?

Bunuel, from an algebraic standpoint, if we manipulate Statement II like below, why does the inequality leave open the possibility that N^2 can be a negative fraction? I get why Neg + Neg < 0, but was wondering about the below. Thank you.

N+(1/N)<0 -->
N<-(1/N)
N^2>-1

Not following you... We are asked to find which of the options must be true while given that N is negative (negative integer, negative fraction, negative irrational number). For negative N, N +1/N < 0 must be true. Can you please elaborate what you mean? Thank you.

Sry, my question was more of a general one. Assuming n is neg, I was playing around with the inequality to see if I could manipulate it to coincide with what was already quite apparent (negative + negative = negative).

Basically, I subtracted the negative fraction to the opposite side of the inequality and then multiplied the denominator to the original side (flipping the inequality in the process with N neg). I ended up with n^2 which is is presumed to be positive. Everything seemingly checks out as the inequality says n^2 is > -1, however this includes >=0 n^2 >-1 as well, which seems erroneous.

I was just wondering how to interpret this and if I am making any missteps in my algebraic manipulations and/or thought process.

Thanks very much for your help/insight.

I guess you want to solve for which range of n, n+1/n<0 holds true...

\(n+\frac{1}{n}<0\) --> \(\frac{n^2+1}{n}=\frac{positive}{n}<0\) --> positive/n to be negative, n must be negative, thus \(n+\frac{1}{n}<0\) holds true for \(n<0\).

Hope it helps.

Option E.
The first statement:N^3N^3=-ve
And N^2=+ve since square is always +ve

The second statement is also true because N+1/N=(N^2+1)/N=-ve since N^2+1 will be +ve and N is given -ve.

No need to look at Statement 3 since no option says all three correct.

Posted from my mobile device

Yes it does, much appreciated

[quote="Anasarah"]If N is a negative, which of the following must be true?

I. \(N^3<N^2\) ===================> This will always be true as cube of a negative number will be negative, and square will be positive. => TRUE
II. \(N+\frac{1}{N}<0\) =======> This will again always remain true as we are adding two negative numbers, which will always be < 0 => TRUE
III. \(N=\sqrt{N^2}\) ====> As N is a negative number, and square root can give us both positive and negative numbers, this is not true => FALSE

Hence, the answer is I & II are True - which is option E

I. \(N^3<N^2\)
True as negative<positive.

II. \(N+\frac{1}{N}<0\)
Always true as negative+ negative is less than 0.

III. \(N=\sqrt{N^2}\)
Not true as square root is positive.

Hence, OA is E.