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.