# Why does it make sense to define sqrt(-1) = i but not to define i = sqrt(-1) ?

Sep 30, 2016

As far as I knot, the definition of $i$ is not the one you proposed. I saw your bio, and you surely know the story: $i$ is defined as the solution of the equation ${x}^{2} + 1 = 0$. Since this equation is not solvable in $\setminus m a t h \boldsymbol{R}$, we have that the quotient

$\frac{\setminus m a t h \boldsymbol{R} \left[x\right]}{\left({x}^{2} + 1\right)}$

is a field, which is isomorph to $\setminus m a t h {\boldsymbol{R}}^{2}$, since it is its image through $\setminus \phi : a + b x \setminus \to \left(a , b\right)$. So, $i$ is the image of $x$ over $\setminus \phi$, and in this sense we finally define

$\setminus m a t h \boldsymbol{C} = \frac{\setminus m a t h \boldsymbol{R} \left[x\right]}{\left({x}^{2} + 1\right)} = \setminus m a t h \boldsymbol{R} \left[i\right]$.

So, I would personally say that both the definition you propose are at least sloppy, but let's say why one feels "worst" than the other, at least in my opinion: when you define something new, you need to define the new thing in terms of already existing ones. So, defining $\sqrt{- 1} = i$ "feels like" you already have $i$ in your arsenal, and in that context you can speak about square roots of negative numbers. So, it's not a precise definition, but rather something you "accept" because the computations agree. I think it's like when we say that $\frac{1}{\setminus} \infty = 0$. This of course cannot be given by definition, but we can accept it in that sense.

On the other hand, if we define $i$ as $\sqrt{- 1}$, we are defining something new in terms of something that doesn't exists already, and so our definition has no solid ground to stand on.

Recalling: if we ALREADY have $i$ and all the complex number theory in our hand, we can "define" the square root of $- 1$, since it's not a pure definition, but rather a shortcut.

If we need to define $i$ at the very beginning, we need an existing object to identify it: in this sense, defining $i$ as $\sqrt{- 1}$ would sound as a circular argument, since you're supposed to confirm this equality with later computations, which you are assuming since the beginning.