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Re: M23-05 [#permalink]
How will -1 ^ -2 satisfy option 1?

Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider \(B = -2\), \(A = 2\) (the answer is "yes") and \(B = -1\), \(A = -2\) (the answer is "no").

Answer: E
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Re: M23-05 [#permalink]
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davidbeckham wrote:
How will -1 ^ -2 satisfy option 1?

Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider \(B = -2\), \(A = 2\) (the answer is "yes") and \(B = -1\), \(A = -2\) (the answer is "no").

Answer: E


(-1)^(-2) = (-1)^even = 1.

Or: (-1)^(-2) = 1/(-1)^2 = 1/1 = 1.
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Re: M23-05 [#permalink]
So -negative^even will always give positive despite even power being negative?
Bunuel wrote:
davidbeckham wrote:
How will -1 ^ -2 satisfy option 1?

Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider \(B = -2\), \(A = 2\) (the answer is "yes") and \(B = -1\), \(A = -2\) (the answer is "no").

Answer: E


(-1)^(-2) = (-1)^even = 1.

Or: (-1)^(-2) = 1/(-1)^2 = 1/1 = 1.
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Re: M23-05 [#permalink]
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davidbeckham wrote:
So -negative^even will always give positive despite even power being negative?
Bunuel wrote:
davidbeckham wrote:
How will -1 ^ -2 satisfy option 1?

Bunuel wrote:
Official Solution:

Statements (1) and (2) combined are insufficient. Consider \(B = -2\), \(A = 2\) (the answer is "yes") and \(B = -1\), \(A = -2\) (the answer is "no").

Answer: E


(-1)^(-2) = (-1)^even = 1.

Or: (-1)^(-2) = 1/(-1)^2 = 1/1 = 1.


A negative number in even power is always positive.
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Re: M23-05 [#permalink]
can we take A=0?

Combining (1) + (2)
\((-1)^0 = 1\), so\( 0^{-1}=0 \)and not greater than 0
\((-1)^2 = 1\), so \(2^{-1}=0.5\) and is greater than 0

Answer: E
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Re: M23-05 [#permalink]
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AnirudhaS wrote:
can we take A=0?

Combining (1) + (2)
\((-1)^0 = 1\), so\( 0^{-1}=0 \)and not greater than 0
\((-1)^2 = 1\), so \(2^{-1}=0.5\) and is greater than 0

Answer: E


0^(-1) = 1/0^1 = 1/0. We cannot divide by 0, so 0^(-1) is undefined, not 0. 0 to any negative power is undefined.
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Re: M23-05 [#permalink]
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AnirudhaS wrote:
\((-1)^0 = 1\), so\( 0^{-1}=0 \)and not greater than 0


Be careful here: 0^(-1) is not equal to zero. It's undefined (since it "equals" 1/0).

In theory, it should be fine to use zero in any question worded as this one is (and I very often would use zero, if I was picking numbers, because it's such a convenient number to calculate with). But on the real GMAT, they're very careful to ensure that you will always deal with well-defined expressions. So if a question is going to include a fraction like 1/(k-2), say, the question will always tell you that k is not equal to 2, so that you can never even legitimately consider a value of k that makes the fraction undefined. This question really should also restrict the possible values of its unknowns, because otherwise you can arrive at undefined values, and it's not clear how to answer an inequality question about something undefined.
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Re: M23-05 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M23-05 [#permalink]
Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.


Hi Bunuel, I think there is an error in your explanation.

In explanation for (1), you mentioned that when b is negative, for a^b to be positive, a must be an even number, whether negative or positive. That statement is wrong. For a^b to be positive when b is negative, a needs to be positive; it does not need to be an even number. However, if you are trying to further explain what values a can take for b^a to be positive given that b is negative, then yes a just needs to be even regardless of polarity. If you read on, your examples on values for a and b do not show that a^b is positive.

And I believe the concluding statement for (1) is insufficient, not 'sufficient' as currently indicated.
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Re: M23-05 [#permalink]
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hazeljj wrote:
Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.


Hi Bunuel, I think there is an error in your explanation.

In explanation for (1), you mentioned that when b is negative, for a^b to be positive, a must be an even number, whether negative or positive. That statement is wrong. For a^b to be positive when b is negative, a needs to be positive; it does not need to be an even number. However, if you are trying to further explain what values a can take for b^a to be positive given that b is negative, then yes a just needs to be even regardless of polarity. If you read on, your examples on values for a and b do not show that a^b is positive.

And I believe the concluding statement for (1) is insufficient, not 'sufficient' as currently indicated.


There were two typos:
1. Instead of b^a it should have been a^b in the explanation of (1).
2. Instead of "Now sufficient" it should have been "Not sufficient".
Edited. Thank you!
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