This question already has an answer here:

Here are steps of the "proof":


$\Rightarrow 1=\sqrt{1}$

$\Rightarrow 1=\sqrt{-1\times-1}$

$\Rightarrow 1=\sqrt{-1}\times\sqrt{-1}$

$\Rightarrow 1=i\times i$

$\Rightarrow 1=-1$

At which step did things go wrong?


marked as duplicate by Community Jul 8 '15 at 17:17

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    $\begingroup$ (ab)^0.5=a^0.5*b^0.5 only when a,b>0 $\endgroup$ – Brian Cheung Jul 8 '15 at 16:43
  • $\begingroup$ @user3313320 what is the source? $\endgroup$ – Ameer Hamza Jul 8 '15 at 16:53

It is no longer true that $$\sqrt{ab}=\sqrt{a}\sqrt{b}$$ when $a,b$ are not positive real numbers.

Additional remark: One must be careful when talking of $\sqrt{\cdot}$ in the realm of complex numbers, because it cannot be defined globally, i.e. there is no continuous function $$f:\Bbb C\to \Bbb C.$$ such that $f(z)^2=z$ for all $z\in\mathbb{C}$. Formally, we define $$\sqrt{z}=e^{\frac{1}{2}\log z}$$ where $\log$ is the complex logarithm defined on some proper simply connected open subset of $\Bbb C$.

  • $\begingroup$ Can you provide the source? $\endgroup$ – Ameer Hamza Jul 8 '15 at 16:51
  • $\begingroup$ @AmeerHamza If you are looking for a proof that $\sqrt{ab}\neq\sqrt{a}\sqrt{b}$ you just found one! If it were true then $-1=i\cdot i=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=1$. There is need to go further. $\endgroup$ – Spenser Jul 8 '15 at 16:59

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