What's the thing with $\sqrt{-1} = i$ What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!
So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?
 A: This is answered quite succinctly in the MathWorld article on Principal Square Root.
A: The proper way to define $i$ is to define the laws for complex addition and multiplication the usual way using ordered pairs of real numbers, and then define $i$ to be $(0,1)$. This distinguishes it from $-i$ which is $(0,-1)$.
One can then set the root of $-1$ to be $i$, but of course this last step is pretty arbitrary. You can see it either as wrong or as not wrong if you want to work with the last step as an arbitrary definition. The important thing is that we have an unambiguous definition of $i$ using the initial steps specified above.
A: I believe this is a common misconception. Here in Sweden we (or at least I) was taught that $i^2 = -1$, not that $\sqrt{-1} = i$. They are two fundamentally different statements. Of course most of the time one chooses $\sqrt{-1} = i$ as the principal branch of the square root.
A: $\sqrt{-1}$ is an imprecise notation.  There are several ways of making it precise, some of which involve what comes down to an arbitrary choice of square root of $-1$ and some of which don't.  This is because there are several related ways to construct $\mathbb{C}$ from $\mathbb{R}$:


*

*As the ring $F = \mathbb{R}[x]/(x^2 + 1)$ (edit: following in the discussion in the comments, perhaps it would be better to say "an algebraic closure of $\mathbb{R}$.").  The Galois group $\text{Gal}(F/\mathbb{R})$ has order $2$, and complex conjugation $x \mapsto -x$ is its only nontrivial element.  The Galois symmetry here forbids us from distinguishing between $x$ and $-x$.

*As the ring $\mathbb{C} = \mathbb{R}[\imath]/(\imath^2 + 1)$, where we fix the choice $\imath$ of generator.  This is what is typically meant by $\mathbb{C}$.  This field has a distinguished element $\imath$ satisfying $\imath^2 = -1$, and this is what we usually mean by $i$.  $\mathbb{C}$ is isomorphic to $F$, but not canonically so, since $\imath$ can be sent to either $x$ or $-x$ and there is no way to choose between these.

*As the ring $R$ of linear endomorphisms $\mathbb{R}^2 \to \mathbb{R}^2$ of non-negative determinant preserving an inner product $\langle \cdot, \cdot \rangle$.  There are two elements of $R$ squaring to $-1$, corresponding to a rotation and its inverse, and there is no way to choose between them unless $\mathbb{R}^2$ is also equipped with an orientation, which is an identification of its exterior square with $\mathbb{R}$.  This choice of orientation corresponds to the difference between the first and the second constructions above.
But most people will not bother to stop and talk about such subtleties as $F, \mathbb{C}, R$ being non-canonically isomorphic, so they say $\sqrt{-1} = i$ because the truth is (for practical purposes) unnecessarily complicated.
A: Alex it is just a notion. We use the value of $i = \sqrt{-1}$ so that we can work more freely. For example consider the Quadratic Equation $$x^{2}+x+1=0$$ The discriminant for this is $D=b^{2}-4ac=-3$. So the roots have the value $$x = \frac{-1 \pm{\sqrt{3}i}}{2}$$ which looks better when written with an $i$ notation. That's all. 
I don't know how this is taught in the US but to me, i encountered this when i was at high school, learning how to solve for Quadratic Equations when the Discriminant is less than $0$. 
Next, note that $\mathbb{C}$ doesn't have the same ordering as $\mathbb{R}$. That is for any 2 real numbers $a,b$ we have either:


*

*$a>b$

*$a< b$

*$a=b$
But for complex number's this is not true. Since if you take $i$ and $0$, we must have either $i > 0$ or $i < 0$, but this isn't true.
A: This seems to be very little different than the standard notation for $\sqrt{4} = 2$ as opposed to $\sqrt{4} = -2$. By convention, we have a concept of a principal branch of the square root. More importantly, though, is that there is no real difference between letting $i := \sqrt{-1}$ and $-i := \sqrt{-1}$ (though there is an imaginary difference!). One really needs only to notate the two roots of $\sqrt{-1}$ by $\pm i$.
In addition, I first learned about imaginary numbers in a US school and we learned that i was s.t. $i^2 = -1$ as well. As far as I can recall, the first important distinction comes when we consider how to represent the complex plane graphically and through the exponential series - but again, these are merely notational inconveniences rather than big problems.
A: As even the capitalistic Wikipedia acknowledges, there are three laws of dialectics:


*

*The law of the unity and conflict of opposites;

*The law of the passage of quantitative changes into qualitative changes;

*The law of the negation of the negation.


Thus, from law 1, we see that i and -i are both equal and opposite. From law 2, we learn that $\mathbb C$ is qualitatively different from $\mathbb R$. From law 3, we attain the dynamic equilibrium between i and -i that was so powerfully expounded by the revolutionary hero Évariste Galois.  
This at least is how we taught such matters in the USSR.
A: Does $\pm i=\pm\sqrt{-1}$ sound any better ? Also, $(-i)^2=-1$, so I don't see how defining it as $i^2=-1$ would make things any better.
