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Negitive nth roots?

  1. Jul 30, 2009 #1
    This has been bugging me for a few days now. Usually when I have a question about math I can answer it myself, but this, the answer is evading me. I know that the answer to the negitive square root of 100 is .1. How is this number obtained? Is there a graphical representation? Does the number have a deeper meaning? The main reason that I can't find that answer to this is because how can you have negitive 2 numbers? x^-2? I know that this is correct and I get that. But I just don't know how the .1 is obtained.
     
  2. jcsd
  3. Jul 30, 2009 #2

    rock.freak667

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    negative square root means [itex] - \sqrt{number}[/itex].

    What you mean is the inverse square root of 100 or simply [itex](\sqrt{100})^{-1}=0.1[/itex]
     
  4. Jul 30, 2009 #3
    Thank you. But how did they discover that it's the inverse? Or am I not understanding something correctly?
     
  5. Jul 30, 2009 #4
    Just apply the operations. The square root of 100 is 10 and the multiplicative inverse of 10 is 0.1 (that is, the solution to 10x = 1 is x = 0.1). Or are you asking something else..?
     
  6. Jul 30, 2009 #5

    Janus

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    A negative exponent means the inverse of the number.

    e.g.

    [tex]10^{-1}= \frac{1}{10}[/tex]

    Thus

    [tex]10^{-\frac{1}{2}}= \frac{1}{\sqrt{100}} = \frac{1}{10} = 0.1[/tex]

    You can think of it this way:

    if you divide 10^2 by 10 you get 10 which is 10^1, divide by 10 again and you get 1, which is 10^0, divide by 10 again and you get 1/10 which is 10^-1, again and you get 1/100 which is 10^-2, etc.

    The same rule applies to fractional exponents. 100^(1/2) is the square root of of 100 which equals10. Subtracting 1 from the exponent gives you 100^(-1/2) which is the same as dividing by 100, and if you divide 10 by 100 you get 0.1.
     
  7. Jul 30, 2009 #6

    rock.freak667

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    I am not sure if there is a formal proof. If you can't find it, it is probably defined to be that way. 1/an=a-n
     
  8. Jul 30, 2009 #7

    diazona

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    I'm not sure either whether it's a definition or something that can be proven, but in the former case it was defined that way so that the multiplication rule for powers works with negative numbers.
    [tex](a^m)(a^n) = a^{m+n}[/tex]
    so now if you set [itex]m=-n[/itex] you get
    [tex](a^{-n})(a^n) = a^0 = 1[/tex]
    thus
    [tex]a^{-n} = \frac{1}{a^n}[/tex]
     
  9. Jul 31, 2009 #8
    A simple way of thinking it is 10^0/10^1. now, we know that when we divide 2 numbers with the same base (is this the correct word?) i.e. in this case 10, but with different powers, we minus the powers. In this case, if we minus the powers, it equals 10^-1, but if we acctually do the division, it is 1/10. That's how you get it. If you didn't know, 10^0 is 1 because 10^n/10^n. If we minus the powers, n-n=0, so it's 10^0. If we actually do the division, any 2 numbers divided by itself = 1, so 10^n/10^n = 1
     
  10. Jul 31, 2009 #9
    Ooooooh. It is so simple now. Thanks guys.
     
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