M28-21

No, we cannot say that -1/2^(1/3) must be in the set.

I don't quite understand... the problem says if X is in the set, then -X^2 and -X^3 are in the set, but it doesn't say that -X^4 are also in the set ... which is what we are assuming to get from -1/4 to -1/64. Please help - thank you!

The rule states that if a number is in the set, then its negative squared and negative cubed values must also be in the set.

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

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

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

By the same logic, if \(-\frac{1}{4}\) is in the set, then the following must also be in the set:

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

By the same logic, if \(-\frac{1}{8}\) is in the set, then the following must also be in the set:

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

Bunuel Why cant we go backwards.. like say if 1/2 exists in the set then (1/\sqrt{2}) and 1/[cube_root]2[/cube_root] also would be existing for 1/2 to exist in the set?
Pls explain

If x then y does not mean if y then x.

As per me, a set by definition is a collection of finite numbers. If we go by the explanations, the given set can go on to become a collection of infinite numbers. Please correct me if my understanding is flawed.

Yes, this particular set will be infinite. The question asks which of the following MUST be in the set and both \(-\frac{1}{64}\) and \(\frac{1}{64}\) will definitely be in the set, though \(\frac{1}{\sqrt[3]{2}}\) will not necessarily be there.

bunuel,
pls explain how you got 1/64. i understand how you got -1/64.
thanks

Since 1/2 is in the set, then so must be \(-x^2 = -\frac{1}{4}\);

Since \(-\frac{1}{4}\) is in the set, then so must be \(-x^3 = -(-\frac{1}{4})^3=-(-\frac{1}{64})=\frac{1}{64}\).

Hi Bunuel

i understand why i,ii are in the set.

if -1/2 has to be in the set
then \(\frac{1}{\sqrt[3]{2}}\) can also be in the set right?

May be that value is in the set to get -1/2

Please explain this.

Note that the question asks which of the following must be in the set, not could be in the set. While any number theoretically could be in the set, from the given options, we can only be sure that \(-\frac{1}{64}\) and \(\frac{1}{64}\) will definitely be in the set (because we are given that \(\frac{1}{2}\) is in the set).

Also, even if we were told that \(-\frac{1}{2}\) is indeed in the set, we still could not be sure that \(\frac{1}{\sqrt[3]{2}}\) will definitely be in the set. You see, we are told that if \(x\) is in the set, then \(-x^3\) will also be in the set, not vice versa. For example, if we were told that 8 is in the set, it would not necessarily mean that -2 is in the set. On the other hand, if -2 is in the set, then \(-(-2)^3 = 8\) will for sure be in the set.

Hope it's clear.

I've seen similar questions on this site that come to a different answer. There is nothing that says the set goes on infinitely. The question doesn't state the rule extends for all potential values of x.
The set could very well ONLY be the set of 1/2, -1/4, -1/8. The other numbers COULD be in the set, but MUST be in the set isn't really true.

How are we supposed to accurately differentiate between this logic in DS questions?

Here is a GMAT Prep question which is worded exactly the same way: https://gmatclub.com/forum/for-a-certai ... 36580.html

Isn't -1/2^2 = 1/4, not -1/4? It would be different if the negative sign were outside of the parentheses like -(1/2^2), which would equal -1/4. Sorry if I missed something simple.

Thanks!

Posted from my mobile device

________________________________
(-1/2)^2 = 1/4 but -1/2^2 = -1/4.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.