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

1 Answer
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 \mathbb{R}, we have that the quotient

(\mathbb{R}[x])/((x^2+1))

is a field, which is isomorph to \mathbb{R}^2, since it is its image through \phi: a+bx\to (a,b). So, i is the image of x over \phi, and in this sense we finally define

\mathbb{C} = (\mathbb{R}[x])/((x^2+1)) = \mathbb{R}[i].

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 1/\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.