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I^2 = -1 or +1

  1. Dec 13, 2009 #1
    i was taught that i2 = -1
    because ((-1)1/2)2=(-1)2/2=(-1)1=-1

    but isnt it mathematically correct to say
    [tex]\sqrt{-1}[/tex] * [tex]\sqrt{-1}[/tex] = [tex]\sqrt{-1*-1}[/tex] = [tex]\sqrt{1}[/tex] = 1

    what the heck!!
    Is there some bleedingly obvious thing im missing.
     
  2. jcsd
  3. Dec 13, 2009 #2
    i^2=-1

    Since

    i=[tex]\sqrt{-1}[/tex]
     
  4. Dec 13, 2009 #3

    D H

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    No, it's not correct, and you just discovered why. Exponents don't work quite the same way in the complex numbers as they do in the reals. In the reals, that expression isn't valid because the left-hand side involves two numbers that are not real. In the complex numbers, the square root operation does not distribute over multiplication.
     
  5. Dec 13, 2009 #4

    Char. Limit

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    That particular property ([tex]\sqrt{a*b}=\sqrt{a}*\sqrt{b}[/tex]) only holds true if a and b are positive.
     
  6. Dec 14, 2009 #5
    OK, so what mathematical rules can I define to work with exponents and still be on the safe side with logical truth?
     
  7. Dec 14, 2009 #6
    [tex]i[/tex] is constructed as [tex](0,1) \in \mathbb R x \mathbb R[/tex]

    [tex]i^2 = -1[/tex] and therefore that [tex]i = \sqrt{-1}[/tex]
    follows directly from how algebraic operations are defined on that field. the rule you're talking about does not hold in [tex]\mathbb C[/tex]
     
  8. Dec 14, 2009 #7
    logical truth... that is beyond the scope of mathematics. mathematics is a game of formal systems: axioms and rules of inference.

    if you can construct a field where [tex]\sqrt{-1} \sqrt{-1} = 1[/tex] than it's perfectly mathematically valid...
    just as it's mathematically valid that in [tex] \mathbb C[/tex], [tex]i^2= -1[/tex]
     
  9. Dec 14, 2009 #8
    Here's another method that contains a fallacy:
    1 = 1
    (-1)^2 = 1^2
    sqrt (-1)^2 = sqrt 1^2
    -1 = 1
     
  10. Dec 14, 2009 #9
    no.... when you square a square root the number under the radical remains in tact. isquared is not 1s.
     
  11. Dec 14, 2009 #10

    Hepth

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    does line 1 have something to do with line 2?
    and you forget the +- solutions for the square root...
     
  12. Dec 14, 2009 #11

    Mentallic

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    [itex]1=1^2[/itex] and also [itex]1=(-1)^2[/itex] so yes, line 1 does have a little something to do with line 2.

    No, the square root of a number is almost always defined as the positive value. [itex]\sqrt{9}\neq \pm 3[/itex], it only is +3.

    However, when you solve [itex]x^2=9[/itex] then you need both positive and negative values. [itex]x=\pm (\sqrt{9}) =\pm (3) =\pm 3[/itex]
     
  13. Dec 14, 2009 #12

    Hepth

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    Which is what you are doing. You are saying:

    (-1)^2 = (1)^2 which is true
    operate on this as it were algebraic :
    (A)^2 = (1)^2
    A = +- Sqrt[1^2]
    A = +- 1

    Which are the two solutions to the first equation. I think you're not following proper operational methods by ASSERTING that you can raise each side without taking into consideration.
     
  14. Dec 15, 2009 #13
    sqrt(-a)=sqrt[i2a]

    sqrt[i2a]*sqrt[i2b] = (i*i)sqrt[a]sqrt= -(sqrt[ab])
     
  15. Dec 15, 2009 #14

    Mentallic

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    So you've shown that [itex]A=\pm 1[/itex]. Well, yes... substitute A back into it and you get:

    [tex](\pm1)^2=(1)^2[/tex]

    which is what we've already stated. I don't see what you're trying to get at here.
    This is different to what you said earlier about having a [itex]\pm[/itex] solution when you take the square root.
     
  16. Dec 15, 2009 #15

    Fredrik

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    The square root and other non-integer exponents can be defined using the complex logarithm, which is defined as the multivalued "inverse" of the complex exponential. Why multivalued? Note that we have

    [tex]e^z=e^{z+i2\pi n}[/tex]

    for any integer n. What's [itex]\log e^z[/itex] supposed to be? z? It can't be, since that would imply that

    [tex]z=\log e^z=\log e^{z+i2\pi n}=z+i2\pi n[/tex]

    for all n, which implies that all integers are zero.

    We can avoid this problem by defining

    [tex]\log z=\log(|z|e^{i Arg\ z})=\log |z|+i Arg\ z+i2\pi n[/tex]

    The right-hand side should actually be interpreted as the set

    [tex]\{\log |z|+i Arg\ z+i2\pi n|n\in\mathbb Z\}[/tex]

    but it's too awkward to always use that notation. Note that when we take the logarithm of both sides of the first equation in this post, we get the same thing on both sides, so we have at least solved that problem.

    Now we can define [itex]z^a[/itex] by

    [tex]z^a=e^{\log z^a}=e^{a\log z}[/tex].

    The square root is the special case a=1/2.

    [tex]\sqrt z=e^{\frac 1 2 \log z}=e^{\frac 1 2(\log |z|+i Arg\ z+i2\pi n)}=\pm\sqrt{|z|}e^{\frac i 2 Arg\ z}[/tex]

    So with this definition we get [itex]\sqrt{-1}=\pm i[/itex]. If we instead choose to define [itex]\sqrt z[/itex] as the single value [tex]\sqrt{|z|}e^{\frac i 2 Arg\ z}[/tex], we get [itex]\sqrt{-1}=i[/itex].
     
    Last edited: Dec 16, 2009
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