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:

\sqrt{-1} &= i \\\\\
\frac1{\sqrt{-1}} &= \frac1i \\\\
\frac{\sqrt1}{\sqrt{-1}} &= \frac1i \\\\
\sqrt{\frac1{-1}} &= \frac1i \\\\
\sqrt{\frac{-1}1} &= \frac1i \\\\
\sqrt{-1} &= \frac1i \\\\
i &= \frac1i \\\\
i^2 &= 1 \\\\
-1 &= 1 \quad !!?


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 $b>0$.

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 $b>0$. 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 ($x<0$) 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.

Source : Link , Question Author : Wilhelm , Answer Author : Community

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