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My physical chemistry textbook seems to be making the implicit assumption that $\cfrac{1}{i} = -i$.

I'm not sure how this is valid. Here is the snippet of relevant steps:

$\cfrac{i\hbar}{f(t)}$$\cfrac{df(t)}{dt}$$ = E$ ---I can see multiplying by $f(t)$ here and dividing by $i\hbar$

$\cfrac{df(t)}{dt} = $$ -\cfrac{i}{\hbar}$$Ef(t)$

[If you're wondering,this is part of the time-dependent Schrodinger equation.]

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    $\begingroup$ $i^{2}=i\circ i=-1\Rightarrow i=\frac{-1}{i}$ $\endgroup$ Jun 23 '15 at 22:40
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    $\begingroup$ Depending on your knowledge of algebra this is also a consequence of the uniqueness of inverses in a group... $\endgroup$
    – hjhjhj57
    Jun 23 '15 at 22:42
  • $\begingroup$ I just found that squaring both sides yields $-1 = -1$, too...yeahp. $\endgroup$
    – khaverim
    Jun 23 '15 at 22:47
  • $\begingroup$ The command \hbar in math mode produces $\hbar$. $\endgroup$ Jun 24 '15 at 9:37
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    $\begingroup$ @khaverim Be careful. Squaring both sides of $i = -i$ also yields $-1 = -1$ but this doesn't mean that $i = -i$ is true. $\endgroup$
    – Zoe H
    Jun 24 '15 at 10:42
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$$ \frac{1}{i} = \frac{1}{i}\cdot\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1}==-i $$

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  • $\begingroup$ ah..pretty simple. I forgot about this, thank you $\endgroup$
    – khaverim
    Jun 23 '15 at 22:43
  • $\begingroup$ @khaverim glad you got it now. $\endgroup$
    – Chinny84
    Jun 23 '15 at 22:44
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Another proof is that for any number $n$ its inverse is the number $\frac 1n $ such that $n \times \frac 1n=1$ and we have that:

$$i\times -i=-(i\times i)=-(-1)=1$$

So we can conclude that:

$$\frac 1i=-i$$

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    $\begingroup$ ...because inverses in a group are unique. $\endgroup$ Jun 24 '15 at 10:40

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