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I have created a proof that$ 1 = -1$ but I know that this is impossible. Could someone help me find the flaw in this proof...

$i = \sqrt{-1}$

Given

$i^2 = -1$

Given

$i^4 = 1$

Given $i^8 = 1$

Given --------------------------All Common Knowledge Above

$i^4= i^8$

Take sqrt of both sides...

$i^2=i^4$

Take sqrt of both sides...

$i= i^2$

$i=-1$

$i^2= -1$ (sub for $i$)

$-1 \times -1$ (sub for $i^2$) = $1$

$1=-1$

Thankyou all for helping me. I looked at the other questions and this question is not a duplicate. However, we all have one common error; we forgot +- when taking the square root of i^4 = i^2

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    $\begingroup$ $a^2 = b^2$ does not imply $a = b$. $\endgroup$
    – user61527
    Jul 18, 2014 at 6:07
  • $\begingroup$ The question could use better formatting. $\endgroup$
    – user147263
    Jul 18, 2014 at 6:19
  • $\begingroup$ We have fairly good intuition about positive numbers. Negative, not so much. Nonreal, less still. Applying so-called rules of algebra beyond our zone of comfort has to be done very carefully. $\endgroup$ Jul 18, 2014 at 6:23
  • $\begingroup$ Why are there so many downvotes. Seems like valid question $\endgroup$ Jul 18, 2014 at 7:56
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    $\begingroup$ @AsafKaragila I don't think it's a duplicate of the one you linked to, which is about getting $\sqrt{ab} = \sqrt{a} \sqrt b$ wrong. I'm sure this is a duplicate of something, though. $\endgroup$
    – user61527
    Jul 18, 2014 at 8:14

2 Answers 2

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What people seem to be calling you out for without explanation is that if $a^2 = b^2$, then we can have that $a = \pm b$. We can't know which of $b$ or $-b$ we started with, though.

So when you say that $i^4 = i^8$, then good. You're on the right track. But your next step needs to be that $i^2 = \pm i^4$. You then have no contradiction because one of $i^4$ and $-i^4$ is certainly equal to $i^2$.

I hope this helps you. Always try things like these, even if others call you out for being silly. Before being amazing you have to be a little silly.

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    $\begingroup$ +1 for "Before being amazing you have to be a little silly." $\endgroup$
    – Vincent
    Jul 18, 2014 at 8:08
  • $\begingroup$ Ah, thank you so much. The irony is that none of my highschool math teachers knew why it is wrong either. Do you know who else made this famous mistake? Einstein. E = +-MC^2 which demonstrates the existence of negative energy. $\endgroup$ Jul 18, 2014 at 14:15
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This is a typical example of the misuse of the "$\sqrt .$" function. You said that $\sqrt{i^4}=i^2$ which is not true since $i^4=1$ and $\sqrt{1}=1$, not $i^2=-1$.

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    $\begingroup$ Can anyonw show me what I did wrong instead of just downvoting me? Does anyone think that $\sqrt{i^4}=i^2$? $\endgroup$
    – 5xum
    Jul 18, 2014 at 6:26
  • $\begingroup$ I did not downvote, but $\sqrt{i^4} = i^2$ is certainly possible since $i^2 \cdot i^2 = i^4$. The "true" answer would be that $\sqrt{i^4} = \pm i^2$, as the others have pointed out. $\endgroup$
    – naslundx
    Jul 18, 2014 at 8:01
  • $\begingroup$ @naslundx That is completely false. The square root function is a function mapping positive reals to positive reals. For example, $\sqrt 4 = 2$. The point is that from $x^2=y$, if $y>0$, we can only conclude that $x=\sqrt y$ OR $x=-\sqrt y$. However, $\sqrt y$ is still a well defined number that is not one of two numbers. $\endgroup$
    – 5xum
    Jul 18, 2014 at 8:04
  • $\begingroup$ There ought to be $i^2$ in the last phrase, I believe. $\endgroup$
    – user61527
    Jul 18, 2014 at 8:15
  • $\begingroup$ @T.Bongers Thank you. Fixed. $\endgroup$
    – 5xum
    Jul 18, 2014 at 8:17

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