Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$?

I recognize there have to be something unmathematical in the adhering to yet I do not recognize where it is:

\begin{align} \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 !!! \end{align}

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2019-05-04 17:28:38
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Answers: 2

Between your 3rd and also 4th lines, you make use of $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$. This is just (assured to be) real when $a\ge 0$ and also $b>0$.

modify : As mentioned in the remarks, what I suggested was that the identification $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$ has domain name $a\ge 0$ and also $b>0$. Outside that domain name, using the identification is unacceptable, whether it "jobs."

As a whole (and also this is the core of the majority of "phony" evidence entailing square origins of adverse numbers), $\sqrt{x}$ where $x$ is an adverse actual number ($x<0$) have to first be revised as $i\sqrt{|x|}$ prior to any kind of various other algebraic adjustments can be used (due to the fact that the identifications connecting to adjustment of square origins [probably exponentiation with non - integer backers as a whole ] call for nonnegative numbers).

This similar question, concentrated on $-1=i^2=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}\overset{!}{=}\sqrt{-1\cdot-1}=\sqrt{1}=1$, is making use of the comparable identification $\sqrt{a}\sqrt{b}=\sqrt{ab}$, which has domain name $a\ge 0$ and also $b\ge 0$, so using it when $a=b=-1$ is void.

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2019-05-08 05:20:14
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Isaac's answer is proper, yet it can be tough to see if you do not have a solid expertise of your regulations. These troubles are usually very easy to address if you analyze it line by line and also streamline both sides.

$$\begin{align*} \sqrt{-1} &= \hat\imath & \mathrm{LHS}&=i, \mathrm{RHS}=i \\ 1/\sqrt{-1} &= 1/\hat\imath & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \sqrt{1}/\sqrt{-1} &= 1/\hat\imath & \mathrm{LHS}&=1/i=-i, \mathrm{RHS}=-i \\ \textstyle\sqrt{1/-1} &= 1/\hat\imath & \mathrm{LHS}&=\sqrt{-1}=i, \mathrm{RHS}=-i \end{align*}$$

We can after that see that the mistake has to be thinking $\textstyle\sqrt{1}/\sqrt{-1}=\sqrt{1/-1}$.

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2019-05-08 05:16:06
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