0

I know that ${1\over i} = -i$ where $i = \sqrt{-1}$. Can anyone figure out the mistake in the following lines?

1/i = 1/squareroot(-1) = squareroot(1)/squareroot(-1) = squareroot(1/(-1)) = squareroot(-1/1) = squareroot(-1) = i

${1\over i}={1\over \sqrt{-1}}={\sqrt{1}\over \sqrt{-1}}=\sqrt{1\over -1}=\sqrt{-1\over 1}=\sqrt{-1}=i$

Which is a contradiction? I tried hard, but I am not able to figure out what's wrong with the above equations. Thank You.

gymbvghjkgkjkhgfkl
  • 8,090
OzoneX
  • 3

3 Answers3

2

You simply found one of the instances where, after defining $\sqrt\bullet:\Bbb C\to \Bbb C$ to be the principal value of the square root, you must face the fact that some identities you were familiar with for the usual square root $[0,\infty)\to[0,\infty)$ no longer hold: it is not, in this generality, the case that $\sqrt{a}\sqrt b=\sqrt{ab}$, or that $\sqrt{\frac ab}=\frac{\sqrt{a}}{\sqrt b}$, or even that $\sqrt{\frac1b}=\frac{1}{\sqrt b}$.

  • 1
    "You simply found one of the instances where, after defining $\sqrt{\bullet}:\mathbb C\to\mathbb C$ to be "the only continuous function $f$ that satisfies $f^2=id$ and such that $f(−1)=i$"" — If you define $\sqrt{\bullet}$ like this, you will find it undefined. There is no continuous function with these properties. – celtschk Mar 04 '17 at 18:16
  • @celtischk True. That was a very poor (very wrong) wording of what a had in mind. –  Mar 04 '17 at 18:48
  • Are there any general rules for the indices of complex numbers? If yes can you share some source. Thank you. – OzoneX Mar 04 '17 at 19:52
1

$$\frac 1i = \frac 1i \cdot \frac{-i}{-i} = \frac {-i}{-i^2} = \frac {-i}{-(-1)} = \frac {-i}{1} = -i$$

hatim muhammed
  • 67
0

If $a,b \geq 0$ (and $b \neq 0$) then we have that $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Now tell me: are both $1, -1\geq 0$?

Student
  • 4,438