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-1 is not 1, so where is the mistake?
$i^2$ why is it $-1$ when you can show it is $1$?


$$ \begin{align} 1+1 &= 1 + \sqrt{1} \\ &= 1 + \sqrt{1 \times 1} \\ &= 1 + \sqrt{-1 \times -1} \\ &= 1 + \sqrt{-1} \times \sqrt{-1} \\ &= 1 + i \times i \\ &= 1 + (-1) \\ &= 1 - 1\\ &= 0 \end{align} $$

I can't see anything wrong there, and I can't see anything wrong in $1+1=2$ too. Clearly, $1+1$ is $2$, but I really want to know where is the incorrect part in the above.

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marked as duplicate by Hans Lundmark, Arturo Magidin, Marvis, Zev Chonoles Jun 14 '12 at 20:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You cannot do this operation with complex numbers! – Argon Jun 14 '12 at 20:14
As has been pointed out the error lies in the step that asserts that $\sqrt{(-1)(-1)} = (\sqrt{-1})(\sqrt{-1})$. This identity does not hold for nonnegative factors. Perhaps someone can find the duplicates. – Arturo Magidin Jun 14 '12 at 20:14
Perhaps it was "for non-positive or zero factors..."? – DonAntonio Jun 14 '12 at 21:22
Note that 1+1=0 mod 2, in which case every equality in the post can be interpreted as a true statement (we can take $i=1$ in this context). This shows that the error in the argument is not intrinsic to the algebra, but depends on properties of the ground field/context in which the equations are set. – Mark Bennet Jun 14 '12 at 22:12

3 Answers 3

up vote 7 down vote accepted

$$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$$ is valid only for non-negative real numbers $a$ and $b$. Hence, the error is in the step $$\sqrt{(-1) \times (-1)} = \sqrt{-1} \times \sqrt{-1}$$

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Why is it only valid for non-negative numbers? – Derek 朕會功夫 Jun 14 '12 at 20:21
@Derek $\sqrt x$ is by definition the positive square root of $x$ and is only defined for $x\ge0$. The symbol $\sqrt{-1}$ is an abuse of notation in fact. The definition of $i$ is $i^2 = -1$ – user17762 Jun 14 '12 at 20:24
@Derek: because you get a contradiction otherwise? In the complex numbers it's only valid when $a,b$ have arguments in $[0,\pi)$ due to the standard branch cut. (Technically I think exactly one of $a,b$ may also have argument $\pi$, i.e. be on the negative real axis). – anon Jun 14 '12 at 20:25

We have that $\sqrt{-1} \times \sqrt{-1} = (\sqrt{-1})^{2} = -1$ but $\sqrt{-1 \times -1} = \sqrt{1} = 1$. So $\sqrt{-1 \times -1} \neq \sqrt{-1} \times \sqrt{-1}$ which is the error.

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You cannot split the square root term since that is only valid for non-negative numbers.

$$\sqrt{-1 \cdot -1} \ne \sqrt{-1}\cdot \sqrt{-1}$$

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