For a certain set of numbers, if x is in the set, then both -x^2

Question

For a certain set of numbers, if  x  is in the set, then both  -x^2  and  -x^3  are also in the set. If the number  \frac{1}{2}  is in the set, which of the following must also be in the set?

I.  -\frac{1}{64}

II.  \frac{1}{64}

III.  \frac{1}{\sqrt {2}}

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

Bunuel · Accepted Answer

Official Solution:

For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set, which of the following must also be in the set?

I. \(-\frac{1}{64}\)

II. \(\frac{1}{64}\)

III. \(\frac{1}{\sqrt[3]{2}}\)

A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(I\), \(II\) and \(III\)

Since \(\frac{1}{2}\) is in the set, the following must also be in the set:

\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:

\(-x^3 =\frac{1}{64}\)

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:

\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D

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goforsriram · Answer

Ah Thanks ! I stopped with the first iteration! Should have checked for -1/4 and -1/8 !

aliakberza · Answer

Hi  Bunuel

Could you please confirm whether  (\frac{-1}{2})^\frac{1}{3}  would be in the set? Given that  \frac{1}{2}=-x^3  ==>>  x=-(\frac{1}{2})^\frac{1}{3} . Thanks

Bunuel · Answer

No,  -(\frac{1}{2})^\frac{1}{3}  is not necessarily in the set. We are not told that if x is in the set then  -\sqrt {x}  is in the set.

GDT · Answer

VeritasKarishma

I have a doubt in the highlighted part. we are told that if x is there then  -x^3  will also be there. By this if 1/2 is taken as -x^3, then we'll get  (\frac{-1}{2})^\frac{1}{3}  as x

Can you pls clarify?

KarishmaB · Answer

Are you given that if -x^3 is there in the set, then x is also there? No. This means if I know that -8 is in the set, then I cannot say whether 2 is in the set or not. But I can say that -64 and 512 are in the set. We can only go forward with what we know exists. 
Let's say I start building the set and I put 1/2 in it. Then I must put -1/4 and -1/8 in it too. Then I must put -1/16, 1/64, - 1/64 and 1/512 in it too and so on...

kamrulhasan26043 · Answer

If x is 1/2, then (-x)^2 will be (-1/2)*(-1/2)= 1/4
and (-x)^3 will be (-1/2)*(-1/2)*(-1/2) = -1/8

isn't it?

And A set could contain only  , as the question didn't mention any further where it goes, how can I end up with 1/64 or -1/64?

Could you please  Bunuel   ScottTargetTestPrep   VeritasKarishma , clear my thought.

KarishmaB · Answer

The way the set is defined, it HAS to be a set of infinite elements. 
"if x is in the set, then both -x^2 and -x^3 are also in the set."

If 1/4 is in the set, it is a MUST that -1/16 and -1/64 will be in the set. Going further, since -1/16 is in the set, ...

By its very definition, the set will continue to be formed.

ScottTargetTestPrep · Answer

Solution:

Notice that the exact statement in the question is “if x is in the set, then -x^2 and -x^3 are also in the set”. This is different from saying that “the set consists of x, -x^2 and -x^3 where x = 1/2”. In the latter case, the set does indeed consist of only 1/2, -(1/2)^2 = -1/4 and -(1/2)^3 = -1/8. However, the former sentence tells us that for any element of the set, the negative square and the negative cube of that element are also in the set. Taking x to be 1/2, we can conclude that -1/4 is in the set since -1/4 = -(1/2)^2 and -1/8 is in the set since -1/8 = -(1/2)^3. Now, we can let x be -1/4 or -1/8 and apply the same sentence. Letting x = -1/4, we find that -(-1/4)^3 = 1/64 is in the set. Letting x = -1/8, we find that -(-1/8)^2 = -1/64 is in the set. We can actually let x be 1/64 or -1/64 and continue the above process to find even more elements which are necessarily in the set.

Basshead · Answer

If x is in the set, then -x^2 and -x^3 are in the set.

If 1/2 is in the set, then -1/4 and -1/8 are in the set.

If we know -1/4 is in the set, then (1/4)^3 = 1/64 is in the set.

If we know -1/8 is in the set, then -(-1/8)^2 = - 1/64 is in the set.

We don't know for sure if III. 1/2^(1/3) is in the set.

Answer is D.

MHIKER · Answer

\frac{1}{2}  is in the set, so  (-\frac{1}{2})^2 = \frac{1}{4}  is in the set, Then  (-\frac{1}{4})^3=-\frac{1}{64}  is in the set.

\frac{1}{2}  is in the set, (- \frac{1}{2} )^3 is in the set= -\frac{1}{8}  is in the set, Then (- \frac{1}{8})^2=\frac{1}{64}

I  and  II  both are answers

The answer is  D.

bumpbot · Answer

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For a certain set of numbers, if x is in the set, then both -x^2

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