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How would you proove a^-n = 1/a^n?

  1. Jun 4, 2007 #1
    I can't find many proofs of how the equation in the thread title is correct. How would you proove the equation?

    I tried to make one, but I don't know if it's worth mentioning. What do you think about my "proof"?

    a^-n = a^0-n = (a^0)/(a^n) = 1/a^n
  2. jcsd
  3. Jun 4, 2007 #2


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    you assume that a^b a^c = a^(b+c).
  4. Jun 4, 2007 #3
    I'm sorry, could you elaborate a little? Do I assume? Do you assume? What does that equation have to do with this?

    You might notice that I'm a beginner.

    Edit: Oh, I believe I know what you mean: I did not proove anything, I just wrote something obvious that someone else has figured out? Kind of like not thinking on my own?
    Last edited: Jun 4, 2007
  5. Jun 4, 2007 #4
    The problem is this: what does xn mean when n is a negative integer? If you have defined exponentiation in terms of repeated multiplication, then what does it mean to multiply something by itself -3 or 0 times? Once you have defined exponentiation for these (and other) values, there is nothing really to prove.
  6. Jun 4, 2007 #5

    matt grime

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    You assume that you know what x^2, x^3 means etc. Then you notice that (x^a)^b = x^(ab), and other such rules, that make you realize that it is reasonable to extend thee definition from just the positive whole numbers to all rational numbers via x^{1/n} as the n'th root of x, and x^-n as 1/x^n
  7. Jun 4, 2007 #6
    Because by multiplying both side of the equation by a^m, with m>n, it makes sense.

    Doing so on this equation:

    a^-n = 1/a^n

    leads to

    a^(m-n) = a^m/a^n

    which makes sense (already) for m>=n .
  8. Jun 4, 2007 #7
    in other words: isomorphism ...
  9. Jun 4, 2007 #8


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    What does isomorphism have to do with anything? :confused:
  10. Jun 4, 2007 #9
    well, summing exponents is similar to making products
    there is an isomorphism between:

    - the set of positive numbers equipped with multiplication
    - the set of all numbers equipped with the addition

    (sorry for the lose speaking, this is very old for me)
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