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Is it right to write $i = \sqrt{-1}$?

I know more than well that the complex unity $i$ is defined by $i^2 = -1$.

Still, the $\sqrt{()}$ notation generally allows the following manipulation:

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a\cdot b}$

But what makes me frown, when reading $i = \sqrt{(-1)}$, is that the previous trick can be used to prove this equality wrong:

$i = \sqrt{(-1)}$

$i\cdot i = \sqrt{(-1)}\cdot\sqrt{(-1)}$

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

$i^2 = \sqrt{1}$

$-1 = 1$

However, I really often come across people writing or saying, with a lot of confidence, that $i = \sqrt{(-1)}$. Hence my questions:

  • Is it actually okay to write $i = \sqrt{-1}$?
  • Is the reasoning I wrote correct, or am I expecting too much from the mere $\sqrt{()}$notation?

Thanks for your answers.

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You can write $i = \sqrt{-1}$, but you have to take caution with it. If you extend the function $\sqrt{\cdot} \colon [0,\infty) \to \mathbf R$ somehow, say to some function $\sqrt{\cdot} \colon \mathbf R \to \mathbf C$, you cannot expect it to have the same properties as the function you started with, just because you happen to denote it with the same symbol. You correctly saw that any extension of $\sqrt{\cdot} \colon \mathbf R \to \mathbf C$ with $\sqrt{x}^2 = x$ for all $x\in \mathbf R$ will not have the property that $$ \sqrt{a}\sqrt{b}= \sqrt{ab} $$ Just because $$ \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1 \ne 1 = \sqrt{(-1)(-1)} $$ Hence, you are expecting too much. And to prevent you from it, I'd rather not use the notation.

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  • $\begingroup$ Pretty much the answer I was expecting. I will update my question though, to emphasis the fact that this notation is used pretty much everywhere (which is basically the reason why I posted this question). Thanks for your answer! $\endgroup$ – Right Leg Dec 9 '16 at 7:48
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See: Principal Value to understand how the square-root symbol is used by mathematicians. The argument "confusing to beginners" did not seem to convince them to stop using this notation.

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  • $\begingroup$ Yeah, I had read that. Thing is, in English, the term "principal value" or "principal square root" is used, while in French, we define the square root as "the positive x value such that x^2 = ...". Well at least, some of my teachers defined it this way. $\endgroup$ – Right Leg Dec 9 '16 at 16:39
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You are mistaken when you say it generally allows that transformation. In fact, the whole problem is that that transformation is NOT allowed generally, it is only allowed in special cases - to wit, when $a$ and $b$ are both non-negative. [Believe me: historically a LOT of confusion was caused because even some very smart people took a while to grasp that that transformation wasn't universally valid, so don't feel too bad you got caught as well.]

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Trying to define "i" as $\sqrt{-1}$ leads to problems as you suggest. Better is to define the set of all complex numbers as pairs of real numbers, (a, b) with addition "coordinate wise", (a, b)+ (c, d)= (a+ c, b+ d), and multiplication defined as (a, b)(c, d)= (ac- bd, ad+ bc). That satisfies all of the usual arithmetic properties- addition and multiplication are associative, commutative, multiplication distribute over addition, etc. We can think of the real numbers as a subset of the complex numbers by identifying the real number "x" with the pair (x, 0). And we also have (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0). Since (-1, 0) is identified as the real number -1, denoting (0, 1) as "i" gives $i^2= -1$.

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