Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? I know there must be something unmathematical in the following but I don't know where it is:
\begin{align}
\sqrt{-1} &amp;= i \\\\\
\frac1{\sqrt{-1}} &amp;= \frac1i \\\\ 
\frac{\sqrt1}{\sqrt{-1}} &amp;= \frac1i \\\\ 
\sqrt{\frac1{-1}} &amp;= \frac1i \\\\ 
\sqrt{\frac{-1}1} &amp;= \frac1i \\\\ 
\sqrt{-1} &amp;= \frac1i \\\\ 
i &amp;= \frac1i \\\\ 
i^2 &amp;= 1 \\\\ 
-1 &amp;= 1 \quad !!?
\end{align}
Between your third and fourth lines, you use $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$.  This is only (guaranteed to be) true when $a\ge 0$ and $b&gt;0$.

edit:  As pointed out in the comments, what I meant was that the identity $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ has domain $a\ge 0$ and $b&gt;0$.  Outside that domain, applying the identity is inappropriate, whether or not it "works."

In general (and this is the crux of most "fake" proofs involving square roots of negative numbers), $\sqrt{x}$ where $x$ is a negative real number ($x&lt;0$) must first be rewritten as $i\sqrt{|x|}$ before any other algebraic manipulations can be applied (because the identities relating to manipulation of square roots [perhaps exponentiation with non-integer exponents in general] require nonnegative numbers).

This similar question, focused on $-1=i^2=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}\overset{!}{=}\sqrt{-1\cdot-1}=\sqrt{1}=1$, is using the similar identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$, which has domain $a\ge 0$ and $b\ge 0$, so applying it when $a=b=-1$ is invalid.
what you are saying is similar to $1^2=(-1)^2$ so $1=-1$, math does not work that way