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Irrationality of Rational Exponents

  1. Aug 28, 2004 #1
    What's actually the definition of [itex]a^{m/n}[/itex] where m and n are integers and a is any real number? Suppose I define it as the n-th square root of [itex]a^m[/itex]. Wouldn't it be inconsistent with other stuffs?

    What stuffs? For example, [itex]a^1[/itex] is supposed to be a. But 1 = 2/2 and, using my earlier definition, [itex]a^{2/2}=\sqrt{a^2} = |a|[/itex]. Thus if we use my definition, [itex]a^1[/itex] wouldn't be the same as [itex]a^{2/2}[/itex] for a < 0.

    So, what's the definition in use for [itex]a^{m/n}[/itex]?

    Thanks a lot.
     
  2. jcsd
  3. Aug 28, 2004 #2

    arildno

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    Short answer:
    [tex]a^{\frac{m}{n}}=Exp(\frac{m}{n}Log(a))[/tex]
    where:
    [tex]Exp(x)=1+\sum_{i=1}^{\infty}\frac{x^{i}}{i!}[/tex]
    and Log(x) is the inverse of the Exp function

    Edit: This works only for positive numbers "a".
    It's "easy" to create an exponent form for a=0.
    For negative numbers a, I think you'll have to resort to complex numbers definitions of powers, which might involve delicate questions concerning the logaritmic branch you use.
    I'm not familiar with how these issues concerned with negative a's are actually resolved by mathematicians.
     
    Last edited: Aug 28, 2004
  4. Aug 28, 2004 #3

    jcsd

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    I think one thing you've missed argo is that [itex](a^2)^{\frac{1}{2}} = \pm \sqrt{a^2}[/itex]

    If we specifically state that a has only one value then we will know it is either [itex]\sqrt{a^2}[/itex] or [itex]-\sqrt{a^2}[/itex]
     
  5. Aug 28, 2004 #4

    arildno

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    I don't think so, jcsd:
    I am quite certain that [tex]a^{\frac{1}{2}}[/tex] signifies the positive number.
     
  6. Aug 28, 2004 #5

    ahrkron

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    It depends on the context, arildno.

    How would you express the question "what number(s), when squared, produce the value a"?

    In some cases you do want only the positive roots, but some times you are interested in all roots; when you are, it is often true that none of them are purely positive (in the sense that they are complex numbers; e.g., the fifth roots of 1+0.1i).
     
  7. Aug 28, 2004 #6

    arildno

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    I was talking about the convention of using the expression [tex]a^{\frac{1}{2}}[/tex] as a positive number, not every solution of every imaginable equation.
     
  8. Aug 28, 2004 #7

    HallsofIvy

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    Since you said "as a positive number", I take it you are talking about real numbers. In that case, ax for x negative, or fractional, is defined only for a> 0.
     
  9. Aug 28, 2004 #8

    jcsd

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    You're right it is defined as the principal value.
     
  10. Aug 28, 2004 #9
    This matter is nothing but a definition, but I had a text that saw it this way:

    [tex]\sqrt{a^2}=[/tex] absolute value of a if a >0 and =-absolute value of a if a< 0. That is to say, you have to assign that value depending upon the case. Does that make sense?
     
    Last edited: Aug 28, 2004
  11. Aug 28, 2004 #10

    HallsofIvy

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    No, that doesn't make sense. That would be the same as saying,
    "Since 4=(-2)2, then [tex]\sqrt(4)= -2[/tex]" as well as
    "Since 4=(2)2, then [tex]\sqrt(4)= 2[/tex]".

    [tex]\sqrt{4}= 2[/tex] no matter how you GOT the 4!
     
  12. Aug 30, 2004 #11

    arildno

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    Agreed. In my edit to my original reply, I broached the subject of negative a's (in an admittedly inadequate manner), and thought this subject must probably be resolved by treating a as a special case of a complex number.
     
  13. Aug 30, 2004 #12

    HallsofIvy

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    Of course once you start using complex numbers many functions (including fractional powers) become many-valued. Real valued functions on the real numbers are always, by definition, single valued.
     
  14. Aug 30, 2004 #13

    matt grime

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    I'm not sure i understand your semantic distinction that means sqrt is many valued from C to C, but not from R to R. Functions are single valued, though we often refer to things as many-valued functions to indicate that some choice is usually required, though this doesn't require the introduction of complex numbers.
     
  15. Aug 30, 2004 #14
    You could be right, but that book I was referring to was a 9th grade text on Trigomentry, if I remember right. Trig signs are usually assigned according to the quatrant they are in.

    What would you do with this? sin(x)=[tex]\sqrt{1-cos(x)^2}[/tex]
     
    Last edited: Aug 30, 2004
  16. Aug 30, 2004 #15

    krab

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    I think you answered your own question. If the sign depends on the quadrant, then the formula [itex]\sin(x)=\sqrt{1-\cos(x)^2}[/itex] is incomplete; you must multiply by a sign. Certainly the formula is not the same as [itex]\sin^2(x)+\cos^2(x)=1[/itex].
     
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