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Indices proof

  1. May 30, 2006 #1

    Hootenanny

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    Something I have been curious about, but never had the time to think about is, I know that;

    [tex]x^{3/2} = \sqrt{x^{3}}[/tex]

    But I have never seen any proof of this. Does anyone have a good resource or can show me the proof here? It would be much appreciated.

    ~H
     
  2. jcsd
  3. May 30, 2006 #2

    dav2008

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    So are you asking why [tex](a^m)^n=a^{mn}[/tex]?

    A simple explanation would just be to expand the left-hand side:

    [tex](a^m)^n[/tex]
    [tex]=a^m \cdot a^m \cdot...\cdot a^m[/tex] (n-times)
    [tex]=a^{m+m+m+...+m}[/tex] (n times) since [itex]a^m \cdot a^m=a^{m+m}[/itex]
    [tex]=a^{mn}[/tex] since m+m+m+...+m n times is just m times n
     
    Last edited: May 30, 2006
  4. May 30, 2006 #3

    Hootenanny

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    Thank's yeah, I've just got it. Just as I was replying to this I found it in one of my old textbooks, guess I should look through my books more before asking stupid questions.:grumpy: Thank's again.

    ~H
     
  5. May 30, 2006 #4

    dav2008

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  6. May 30, 2006 #5

    HallsofIvy

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    This is not so much a "proof" as an explanation of why we define
    ax in certain ways.

    It is easy to show that, as long as n and m are positive integers, anam= an+m. That's just a matter of counting the number of "a"s being multiplied.
    Similarly, it is easy to show that, as long as n and m are positive integers, (an)[/sup]m[/sup]= anm. Again, that's just a matter of counting the number of "a"s being multiplied.

    Those are very nice formulas! It would help a lot if axay= ax+y and (ax)y= axy for all x and y.

    IF it were true that ana0= an+0, even when the exponent is 0, we must have an+0= an= ana0 and if a is not 0 we can divide by an to get a0= 1 as long as a is not 0.

    Similarly, to guarentee that this formula is true for n negative, we must have ana-n= an-n= a0= 1: in other words that a-n= 1/an. Of course, we can only do that division if an is not 0: in other words if a is not 0.

    If we want (an)m even when m is not a positive, we must have (an)-n= an-n= a0= 1. In other words, we must define a0= 1 as long as a is not 0. There is no way to define 00 that will make anam= an+m for a= 0.

    Similarly, if we want anam= an+m for n or m negative, we must have ana-n= an-n= a0= 1 for all positive integers n: in other words, again dividing by an, a-n= 1/an as long as a is not 0.

    As dav2008 pointed out, in order to have (am)n= amn true even when m and n are not integers, we must have (an)1/n= a1 so that a1/n= [itex]\^n\sqrt{a}[/itex] and then, that am/n= [itex]^n\sqrt{a^n[/itex].

    In order to define ax for x irrational we require that f(x)= ax be continuous.
     
  7. May 30, 2006 #6

    Hootenanny

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    Ahh, makes more sense now. Thanks both of you, it is much appreciated.

    ~H
     
  8. May 31, 2006 #7

    Gokul43201

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    One of us is making a mistake with this bit. Is that a typo ?
     
  9. Jun 1, 2006 #8

    Hootenanny

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    Didn't spot that, surely (an)-n = [itex]a^{-n\cdot n}[/itex]?

    ~H
     
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