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anonymous
 5 years ago
explain why any number (except 0) to the zero power always equals 1
anonymous
 5 years ago
explain why any number (except 0) to the zero power always equals 1

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nikvist
 5 years ago
Best ResponseYou've already chosen the best response.0\[a^0=a^{bb}=a^b\cdot a^{b}=a^b\cdot\frac{1}{a^b}=1\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Because 1 is the neutral element of the multiplicative group of real numbers except 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please use smaller words "nowhereman" i am going to write this on thetest and i cant sound overlly smart

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nikvist's algebraic interpretation is correct.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0The above calculation is quite good already. You could also write: \[a^0 \cdot a^n = a^{0+n} = a^n\] so \[a^0 = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0explain why 0 to the 0 power is not one

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Because it is undefined. For the exponential function 0^x to be continuous it must be 1 but for x^0 to be continuous it must be 0. So you can't define it consistently (e.g. so that all powerrules hold)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or you can go back to nikvist's explanation and realize that to get \[0^0\] you'd have to divide by 0 which isn't defined.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"The choice whether to define 0^0 is based on convenience, not on correctness"

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Well, what is correctness anyway. After all you choose which axioms you rely on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't shoot the messenger :(

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Whos quote was it then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Donald C. Benson, The Moment of Proof : Mathematical Epiphanies. New York Oxford University Press (UK), 1999.
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