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Re: If N is a negative, which of the following must be true? [#permalink]
13 Feb 2014, 09:37

Expert's post

Anasarah wrote:

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

I. \(N^3<N^2\) II. \(N+\frac{1}{N}<0\) III. \(N=\sqrt{N^2}\)

A. I only B. II only C. III only D. I and III only E. I and II only

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

I. \(N^3<N^2\) II. \(N+\frac{1}{N}<0\) III. \(N=\sqrt{N^2}\)

A. I only B. II only C. III only D. I and III only E. I and II only

I. \(N^3<N^2\). Since N is negative, then (N^3=negative) < (N^2=positive). Hence, this one must be true,

II. \(N+\frac{1}{N}<0\). Both N and 1/N are negative, the sum of two negative values is negative. The same here: must be true.

III. \(N=\sqrt{N^2}\). The square root function cannot give negative result: \(\sqrt{some \ expression}\geq{0}\). Negative N cannot equal to positive \(\sqrt{N^2}\). Never true.

Re: If N is a negative, which of the following must be true? [#permalink]
13 Feb 2014, 14:32

Expert's post

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.

Re: If N is a negative, which of the following must be true? [#permalink]
14 Feb 2014, 00:48

Expert's post

m3equals333 wrote:

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. _________________

Re: If N is a negative, which of the following must be true? [#permalink]
14 Feb 2014, 15:38

Expert's post

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. _________________

Re: If N is a negative, which of the following must be true? [#permalink]
17 Feb 2014, 06:57

1

This post received KUDOS

Expert's post

m3equals333 wrote:

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 \(n<0\).

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