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Math question

  1. May 22, 2006 #1
    I always learned this in math, but never really questioned it.

    why does [tex] n^0=1 [/tex]
     
  2. jcsd
  3. May 22, 2006 #2

    LeonhardEuler

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    Alright, first you know [itex]x=x^1[/itex] and [itex]x^1x^1=x^{1+1}=x^2[/itex], right? And in general, [itex]x^ax^b=x^{a+b}[/itex] when a and b are greater than or equal to one. So why not define [itex]x^0[/itex] so that this is true even if a or b is zero? If this is the case then
    [tex]x^0x^a=x^{a+0}=x^a[/tex]
    We have x0xa=xa, so as long as x is not zero we can divide by xa to come out with [itex]x^0=1[/itex].

    More generally, powers are defined so that [itex]x^ax^b=x^{a+b}[/itex] is true even when a and b are not integers and even when they are not real.
     
    Last edited: May 22, 2006
  4. May 22, 2006 #3
    A simple proof:

    Because [tex] (n^a)*(n^b) = n^{a+b}[/tex],
    it can be said that [tex] n^0 [/tex] is equivalent to [tex] n^1 * n^{-1}[/tex].
    [tex] n * (1/n) = 1 [/tex],
    therefore, [tex] n^0 = 1 [/tex]
     
    Last edited: May 22, 2006
  5. May 22, 2006 #4
    OH OH ok I get it, thanks alot guys.
     
  6. May 25, 2006 #5

    HallsofIvy

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    In other words, if we want aman= an+m to be true for 0 as well as positive integer value of m and n, we must define a0= 1.
    It is also true that if we want aman= an+m to be true for negative integer powers, then we must have ana-n= a0= 1. In other words, we must define [itex]a^{-n}= \frac{1}{n}[/itex].

    To see how we must define [itex]a^{\frac{1}{n}}[/itex], look at the law
    (an)m= amn.
     
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