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If a and b are nonzero integers, which of the following must be

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If a and b are nonzero integers, which of the following must be  [#permalink]

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New post Updated on: 20 Jul 2013, 10:37
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If a and b are nonzero integers, which of the following must be negative?

A. \((-a)^{-2b}\)

B. \((-a)^{-3b}\)

C. \(-(a^{-2b})\)

D. \(-(a^{-3b})\)

E. None of these

Originally posted by kingflo on 20 Jul 2013, 10:12.
Last edited by kingflo on 20 Jul 2013, 10:37, edited 2 times in total.
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Re: If a and b are nonzero integers, which of the following must be  [#permalink]

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New post 20 Jul 2013, 10:34
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If a and b are nonzero integers, which of the following must be negative?

A. \((-a)^{-2b}=\frac{1}{(-a)^{2b}}=positive\).

B. \((-a)^{-3b}\) --> may be positive (consider a=1 and b=2) as well as negative (consider a=1 and b=1).

C. \(-(a^{-2b})=-\frac{1}{a^{2b}}=-\frac{1}{positive}=negative\)

D. \(-(a^{-3b})\) --> may be positive (consider a=-1 and b=1) as well as negative (consider a=1 and b=1).

E. None of these

Answer: C.
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Re: If a and b are nonzero integers, which of the following must be  [#permalink]

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New post 17 Jan 2018, 16:33
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kingflo wrote:
If a and b are nonzero integers, which of the following must be negative?

A. \((-a)^{-2b}\)
B. \((-a)^{-3b}\)
C. \(-(a^{-2b})\)
D. \(-(a^{-3b})\)
E. None of these



IMPORTANT CONCEPT
Rule #1: EVEN powers are always greater than or equal to zero.
So, (POSITIVE value)^(EVEN integer) > 0, and (NEGATIVE value)^(EVEN integer) > 0


So, the correct answer here is C. Here's why:


C) –[a^(-2b)]
Since b is an integer, we know that -2b is an EVEN integer.
So, we get: –[a^(EVEN integer)]
By our rule, a^(EVEN integer) is greater than or equal to zero
Since a is a NON-zero integer, we can conclude that a^(EVEN integer) is GREATER THAN zero
In other words, a^(EVEN integer) is POSITIVE
This means that -[a^(EVEN integer)] is NEGATIVE

Answer:

Cheers,
Brent
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Re: If a and b are nonzero integers, which of the following must be  [#permalink]

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Re: If a and b are nonzero integers, which of the following must be   [#permalink] 28 Jan 2019, 18:04
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