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What is wrong with my logic?

  1. Feb 16, 2007 #1
    if [tex]e^{\pi\imath}=-1[/tex] then:

    [tex]-e^{\pi\imath}=1[/tex] and,

    [tex]e^{2\pi\imath}=1[/tex]

    then:

    [tex]-e^{\pi\imath}=e^{2\pi\imath}[/tex]

    [tex]\rightarrow e^{2\pi\imath}+e^{\pi\imath}=0[/tex]

    [tex]\rightarrow (e^{\pi\imath})^2+e^{\pi\imath}=0[/tex]

    [tex]\rightarrow (e^{\pi\imath}+1)e^{\pi\imath}=0[/tex]

    then:

    [tex]e^{\pi\imath}=0[/tex]

    and

    [tex]e^{\pi\imath}+1=0[/tex]

    Can somebody explain this contradiction to me?
     
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2

    D H

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    From the very first line you wrote:
    [tex]e^{\pi i} + 1 = 0[/tex]
    You divided by zero when you went from
    [tex](e^{\pi i} + 1)e^{\pi i} = 0[/tex]
    to
    [tex]e^{\pi i} = 0[/tex]
     
  4. Feb 16, 2007 #3
    I thought if [tex]ab=0[/tex] then you could have two solutions a = 0 and b = 0?
     
  5. Feb 16, 2007 #4

    D H

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    Think again. For what values of [itex]x[/itex] is [itex]0\cdot x = 0[/itex] true?
     
  6. Feb 16, 2007 #5
    Whoops, sorry :blushing:
     
  7. Feb 16, 2007 #6

    cristo

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    You're thinking of something like if ab=0, then either a=0 or b=0. However, in this case, you know that [itex]e^{\pi\imath}+1=0[/itex], and so [itex] (e^{\pi\imath}+1)e^{\pi\imath}=0[/itex] tells us nothing about [itex]e^{\pi i}[/itex]
     
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