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Complex fun

  1. Apr 25, 2005 #1


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    Find the flaw in the following reasoning:

    [tex]-4 = -4[/tex]

    [tex]\frac{-4}{1} = \frac{4}{-1}[/tex]

    [tex]\sqrt{\frac{-4}{1}} = \sqrt{\frac{4}{-1}}[/tex]

    [tex]\frac{\sqrt{-4}}{1} = \frac{\sqrt{4}}{\sqrt{-1}}[/tex]

    [tex]\frac{\sqrt{-1}\sqrt{4}}{1} = \frac{\sqrt{4}}{\sqrt{-1}}[/tex]

    [tex]\frac{2i}{1} = \frac{2}{i}[/tex]

    [tex]2i^2 = 2[/tex]

    [tex]-2 = 2[/tex]

    It's clear that the equality [itex]\frac{2i}{1} = \frac{2}{i}[/itex] is wrong because on the LHS we have a complex number (of the form a + ib) but the RHS is not. But the real flaw is probably higher. Any idea?
  2. jcsd
  3. Apr 25, 2005 #2
    the mistake is in passing from the 3rd to last to the 2nd to last line:

    namely, you multiply both sides by i and cancel

    [tex]\frac{i}{i} \neq 1[/tex]

    it is -1.
  4. Apr 25, 2005 #3


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    How do you show that?

    Also, how do you show that

    1/i = -i ?
    Last edited: Apr 25, 2005
  5. Apr 25, 2005 #4
    The domain of the square root operation must be kept to fractions in which only the numerator can be negative in order for it to be well-defined for negative real numbers, ie. to be explicit, your real number must be in the form [tex] a/b, \ a \in \mathbb{R}, \ b \in \mathbb{R}^+/\{0\}[/tex]. Here's something simpler, that also appears to work, but is wrong for the same reason:

    [tex]i = \sqrt{-1} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} \Longrightarrow 1=-1[/tex]

    The root of the problem is that


    is not valid according to common definitions. In fact, using standard definitions, you get

    [tex]\frac{1}{i} = -i,[/tex]

    instead, since

    [tex]\frac{1}{i} = \frac{1}{i}\cdot \frac{-i}{-i} = \frac{-i}{1} = -i.[/tex]
    Last edited: Apr 25, 2005
  6. Apr 25, 2005 #5
    This is wrong.

    As you get quite quickly from my last post,

    [tex] \frac{i}{i} = i \frac{1}{i} = i(-i) = 1.[/tex]

    As a matter of fact, the whole purpose of this is to make sure (well, in fact we assume it, in order to make the complex numbers into a nicely organized field) that

    [tex] \frac{i}{i} = 1.[/tex]

    The REAL root of the problem, of course, is that every complex number has two square roots, and the square root operator must choose only one of them to give you if it is to be well-defined. We then need to choose which "nice" properties we want the operator to keep, and typically, we choose those I have mentioned above.
    Last edited: Apr 25, 2005
  7. Apr 25, 2005 #6
    It might have something to do with the fact that [(-1)^(1/2)]*[(-1)^(1/2)] does not equal [(-1)*(-1)]^(1/2) which equals 1^(1/2), because we know by definition of i that [(-1)^(1/2)]*[(-1)^(1/2)] should equal -1.

    Sorry for the ugly equations; I don't know how to make the fancy things you guys can do.

    EDIT: Uh, never mind, those posts just before mine made sense...
    Last edited: Apr 25, 2005
  8. Apr 25, 2005 #7
    First off,
    [tex]\frac{1}{i} = -i[/tex]
    To see this, I'd first like to point out:
    is just notation for: The number which is the multiplicative inverse of [itex]z[/itex].
    In otherwords, 1/z is that number which when multiplied by z gives 1, ie-
    [tex]z\frac{1}{z} = 1[/tex]

    That being said
    [tex]i^2 = -1[/tex]

    [tex]i\cdot i = -1[/tex]

    [tex]-1 \cdot i \cdot i = (-1)(-1)[/tex]

    [tex]( -i )(i) = i \cdot (-i) = 1[/tex]

    [tex]\frac{1}{i} = -i[/tex]
    by the above argument
  9. Apr 25, 2005 #8
    That's fine (I posted it already above). But your earlier statement,

    [tex]\frac{i}{i} \neq 1,[/tex]

    is wrong (as you in fact showed in your post directly above this one).
  10. Apr 25, 2005 #9


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    haha ok it's that simple.
  11. Apr 25, 2005 #10


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    The fundamental error is writing [itex]\sqrt{-1}= i[/itex]. -1, like any number has two roots. In the real numbers, we can distinguish between the positive and negative roots. Since the complex numbers cannot be ordered, we cannot do that in general. It is better to define complex numbers as pairs of real numbers (a, b) defining sum (a,b)+ (c,d)= (a+c,b+d) and product (a,b)*(c,d)= (ac- bd,ac+bd). Using that definition, the "paradox" disappears.
  12. Apr 25, 2005 #11
    Heh. I made an error, fixed it, and didn't realize the error. :blushing:

    Thanks for pointing it out so kindly ( honest )
  13. Apr 26, 2005 #12
    As far as I know, x over x cancels out to positive one, not negative :surprised
  14. Apr 26, 2005 #13
    [tex]\frac{x}{x} = \Big \{ \begin{array}{cr} 1 & x \neq 0 \\ \mbox{undefined} & x=0\end{array}, \ x \in \mathbb{C}[/tex]
    Last edited: Apr 26, 2005
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