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Algebra 2 problem: a to the zero

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0 to the 0 power is undefined' date=' but if it could be defined, it would be 1.


Here's the explanation:




I disagree with their assessment of the situation.


The essential reason why it is undefined is that the limit of x^y as (x,y)-> (0,0) may attain any value depending on the path chosen, hence the limit doesn't exist.


There are some circumstances (which they've listed, and done a good job of explaining) where it makes sense to "consider" it as one, but teaching either that it is one or should be one in a more general case will lead to difficulties both with indeterminate forms in l'Hospital's rule and with multivariate functions, where it certainly doesn't always equal 1.


In general, I believe that it should be avoided to teach something young (such as saying that x/0 = infinity) which will need to be untaught later.

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I understand a to the zero power is 1. What happens when a is a negative number? For instance, a to the zero when a is -3? Is the answer +3 or -3?


The key here is figuring out what is the base.


-3^0, the base is 3 which is raised to the 0, then the answer is negated, thus -3^0 = -1


(-3)^0, the base here is -3, raised to the 0, so the answer is 1


5x^0, the base is x, so we have 5(1) = 5.

(5x)^0, the base is 5x, so we have 1.


Some calculators will handle -3^0 correctly, others incorrectly, so it's good to experiment with your model if you're using the calculator.

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a^0 (a to the zero power) is the same thing as:




if you 'subtract' the exponent (hidden 1) from the numerator and the denominator you are left with 1/1. This works as long as a is a non-zero integer (so positive and negative non zero are ok).


Why does a have to be an integer? Doesn't this also work if a is any real number?

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Thanks for the reply.

Are these the correct answers? (I don't get the trick!)


Say the opposite of these words: 1. Always 2. Coming 3. From 4. Take 5. Me 6. Down



1. never

2. going

3. to

4. give

5. you

6. up


I feel really stupid. :confused:

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