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We have $i^2 = -1$ where $i = \sqrt{-1}$. Now consider $$\sqrt{-1}\cdot \sqrt{-1} = \sqrt{(-1)\cdot(-1)} = \sqrt1 = 1$$ Which proves $-1 = 1$. Is there anything wrong with the above manipulation?

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  • $\begingroup$ $\surd a \surd b = \surd ab$ is valid only if $a,b\geq 0$ $\endgroup$ Jan 27, 2016 at 12:39
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    $\begingroup$ No flaw, you've just shown in a few steps that mathematics is nonsense. $\endgroup$ Jan 27, 2016 at 12:39
  • $\begingroup$ I like your comment Thomas.... :P $\endgroup$
    – x2da
    Jan 27, 2016 at 12:41
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    $\begingroup$ Since $-1$ is obviously not equal to $1$, the question isn't "Is there a flaw", but rather "Where is the flaw" $\endgroup$
    – 5xum
    Jan 27, 2016 at 12:42
  • $\begingroup$ @ThomasAndrews Nice comment, but you failed to take Poe's law into consideration... $\endgroup$
    – skyking
    Jan 27, 2016 at 12:57

2 Answers 2

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The equality $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ doesn't hold for $a,b<0$

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You can not use the radical properties like you have done in a passage. The product of the square root of $-1$ is not the square root of the product!

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