What is wrong with my logic?

  • Thread starter jason17349
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  • #1
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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?
 
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Answers and Replies

  • #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]
 
  • #3
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I thought if [tex]ab=0[/tex] then you could have two solutions a = 0 and b = 0?
 
  • #4
D H
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I thought if [tex]ab=0[/tex] then you could have two solutions a = 0 and b = 0?
Think again. For what values of [itex]x[/itex] is [itex]0\cdot x = 0[/itex] true?
 
  • #5
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Whoops, sorry :blushing:
 
  • #6
cristo
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I thought if [tex]ab=0[/tex] then you could have two solutions a = 0 and b = 0?
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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