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With simple algebra it can be seen that by multiplying both sides of the equation, $i = 1/i$ by $i$ that we receive $-1 = 1$, which obviously is not true. I am puzzled, however, because I cannot find the fault in the following logic.

Because $i = \sqrt{-1}$, the original equation

$i = 1/i$

Can also be written as

$\sqrt{-1} = \dfrac{1}{\sqrt{-1}}$

Since $1 \cdot 1 = 1$, then $1 = \sqrt{1}$, therefore

$\sqrt{-1} = \dfrac{\sqrt{1}}{\sqrt{-1}}$

This can be rewritten as

$\sqrt{-1} = \sqrt{\dfrac{1}{-1}} = \sqrt{-1}$

Where is the error in my calculations, and if there are none, then what is the correct answer? Thank you

N. F. Taussig
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Kilian Olen
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    Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Sep 20 '21 at 20:51
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    The equality $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for postive real numbers. You cannot go from $\sqrt{1}/\sqrt{-1}$ to $\sqrt{1/-1}$. This is the same as the non-valid "proof" that $1=-1$: $1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = ii = i^2=-1$. – Arturo Magidin Sep 20 '21 at 20:52
  • @ArturoMagidin That sounds like an answer there. – Sandejo Sep 20 '21 at 20:54
  • There are two issues. (A) your argument is upside down and you are really saying $\sqrt{-1}=\sqrt{\frac1{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}= \frac{1}{\sqrt{-1}}$. (B) the second equality is dubious in complex arithmetic and it depends how you define the square-root function – Henry Sep 20 '21 at 20:55
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    Writing $i=\sqrt{-1}$ is one of the worst ideas one could have. See this. – TheSilverDoe Sep 20 '21 at 20:59
  • In addition to Arturo’s comment, note that you can fix your calculation by considering square root as a multivalued function, i.e. $\sqrt 1={\pm1}$, $\sqrt {-1}={\pm i }$. – Milten Sep 20 '21 at 21:01
  • @Sandejo: It's a repeat (finally found a target), and as such, should not be answered, but rather closed as a duplicate. – Arturo Magidin Sep 20 '21 at 21:01
  • It is clear that $\sqrt{\dfrac1{-1}}=\sqrt{-1}=i$. But $\dfrac{\sqrt1}{\sqrt{-1}}$ should be evaluated by multiplying by the conjugate of the denominator (to make it real), giving $\dfrac{\sqrt1}{\sqrt{-1}}=\dfrac{\sqrt1(-\sqrt{-1})}{\sqrt{-1}(-\sqrt{-1})}=-\dfrac i{i(-i)}=-\dfrac i1$. –  Sep 20 '21 at 21:03
  • @ArturoMagidin For being a little more rigorous we can say equality holds only for non-negative real numbers.. – lone student Sep 20 '21 at 21:17
  • Mistake 1: $i$ can not be rewritten as $\sqrt{-1}$. We have $i^2=-1$ but we also have $(-i)^2=-1$ but we have defined what $\sqrt{-1}$ means and whether it applies to $i$ or $-i$. Mistake 2: $1\cdot 1=1$ does not imply $1 = \sqrt 1$. We also have $(-1)(-1)=1$ and just knowing $1^2 =1$ is not enough to say $\sqrt 1=1$. We have to distinctly point out that we define it as $1$ and not as $-1$. Mistake 3: We have no reason to believe that $\frac {\sqrt a}{\sqrt b} = \sqrt{\frac ab}$. There are two possible values for each $\sqrt{}$ term and we have no idea which ones we are applying. – fleablood Sep 21 '21 at 00:08
  • @fleablood Yes, my school book says, I must write $i^2=-1$. But, I remember that, in the site high rep( gold badge user in calculus, real-complex analysis)said me, this is correct and nice to write $i=\sqrt {-1}$... – lone student Sep 21 '21 at 03:17
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    @lonestudent TheSilverDoe may have been overstating it when they say "Writing i=−1−−−√ is one of the worst ideas one could have" But, seriously, it is a really badly misleading statement. Anyone told you that it is "correct" and "nice" I advise you to forget it immediately. Doing so seems to nearly ALWAYS lead to these irritating and frustrating "$1 = \sqrt{1} = \sqrt{(-1)\times (-1)}=\sqrt{-1}\times \sqrt{-1} = i\times i = i^2 = -1$: Huh?" errors that should just be avoiding by noting ALL non-zero numbers have two square roots and we can't do this stuff as though there is only one. – fleablood Sep 21 '21 at 04:24
  • I'm seconding @TheSilverDoe's linke to Wikipedia. This explains this very clearly and succinctly and does this exact question. – fleablood Sep 21 '21 at 04:29

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