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Please help me with the 3rd point of this PS prob.

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Please help me with the 3rd point of this PS prob.

by aman88 » Tue Dec 11, 2012 2:42 am
Q: If k=-1, which of the following is (are) true?

1. k^k=k
2. | k| =-k
3. k^0=-k

A. 1 only
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only
E. 1. 2. and 3.

Can someone please explain me the third point (k^0=-k) of this question? How is -1^0=-1?
My answer was B but OA E.

Thanks.
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by GMATGuruNY » Tue Dec 11, 2012 4:00 am
aman88 wrote:Q: If k=-1, which of the following is (are) true?

1. k^k=k
2. | k| =-k
3. k^0=-k

A. 1 only
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only
E. 1. 2. and 3.

Can someone please explain me the third point (k^0=-k) of this question? How is -1^0=-1?
My answer was B but OA E.

Thanks.
1. k^k=k
(-1)^(-1) = -1
1/(-1)^1 = -1
1/(-1) = -1
-1 = -1.
Since statement 1 is true, eliminate D.

2. |k| =-k
|-1| = -(-1)
1 = 1.
Since statement 2 is true, eliminate A and C.

3. k^0=-k
(-1)^0 = -(-1)
1 = 1.
Since statement 3 is true, eliminate B.

The correct answer is E.
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by aman88 » Tue Dec 11, 2012 4:26 am
Thanks Mitch. :)

I am sorry to ask a silly question but, uhh, how is -1^0=1? Is it a rule? Any integer^0 = 1??
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by The Iceman » Tue Dec 11, 2012 6:27 am
aman88 wrote:Thanks Mitch. :)

I am sorry to ask a silly question but, uhh, how is -1^0=1? Is it a rule? Any integer^0 = 1??
Yes, but only if the integer is non-zero.

The value that 0^0 takes has been long argued. Many mathematicians feel that 0^0 should equate to 1. However there is also a school of thought that says 0^0 is indeterminate, or undefined.

Just one line of thought -> k^0 = k^(k-k)=k^k/k^k, which is not defined for k=0 [since, 0/0 is not defined]
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