Not defining the imaginary number $i$ as the principal square root of $-1$.

Question

Background

I learned early on that it's important that we define the imaginary number $i$ such that $i^2 = -1$, rather than $i = \sqrt{-1}$.

Question

I can't fully remember the reasoning for this important note, so I was wondering if someone could elaborate?

Own efforts

Any explanation I find eventually boils down to the same argument.

If we define $i$ as the principal square root of $-1$, then we get

$$-1 = i^2 = \sqrt{-1}\sqrt{-1} \overbrace{=}^{\text{fallacy}} \sqrt{(-1)(-1)} = \sqrt1 = 1$$

But to me, this seems like wrongful use of the $\sqrt{ab} = \sqrt a \sqrt b$ rule, since this rule comes with certain restrictions on $a, b$. So I don't see how this is a misuse of the definition of $i$.

Are there other reasons why we should be careful not to define $i$ as the principal square root of $-1$?

Answer 1

If you define $i$ as $\sqrt{-1}$ then there is an obvious question: how do you know that $-1$ has some square root? Besides, writing $i=\sqrt{-1}$ seems to imply that $i$ is the square root of $-1$. But, in $\mathbb C$, $-1$ has two square roots: $\pm i$. Assuming that $i$ is the square root of $-1$ leads to fallacies, such as the one the you mentioned.

Answer 2

There are obviously two square roots of $-1$ (I'll assume there is a square-root of $-1$, such that that constructed from field extension), let's write them as $\pm\sqrt{-1}_{me}$. And each person, knowing only arithmetics on $\mathbb{R}$, is free to choose his/her $\sqrt{-1}_{person}\in\{\pm\sqrt{-1}_{me}\}$, so there is no guarantee that $\sqrt{-1}_{me}=\sqrt{-1}_{you}$. This is rather undesirable, as we never actually managed to pin down exactly what we meant by $i=\sqrt{-1}$.

Answer 3

Well, one can define the complex numbers as ${\Bbb C} = \{a+ib\mid {\Bbb R}\}$, where $a+ib$ are considered as formal sums. Then one only needs that $i^2=-1$ for defining the multiplication.

The addition is componentwise, $$(a+ib) + (c+id) = (a+c) + i(b+d)$$ and the multiplication (by multiplying term by term) is $$(a+ib)\cdot (c+id) = ac + iad + ibc + i^2bc = (ac-bd) + i(ad+bc),$$ where $i^2=-1$ is used but not the square root.

Moreover, algebraically, $i^2=-1$ implies that $i$ is a primitive fourth root of unity, and, graphically, the numbers $a+ib$ can be drawn as usual in the Gaussian plane. No need of the square root here.

Answer 4

I am not an expert, but I think that if they first declared $\sqrt{-1}=i$, it would not immediately imply that $i^2=-1$, because it could mean that the principal square root operator has one behavior on one domain and another behavior on another. To specify that the first behavior in particular extends to the negative numbers is just to say that $i^2=-1$ anyway (removing the restriction that the principal square root is $\ge0$), so it is more straightforward to just say $i^2=-1$.

Also, there is no way to order the complex numbers for $\ge$ to be applied to all of them, so it is hard to obtain a variation of the $\ge0$ rule to select which root is the principal root. By convention, the principal square root of $-1$ is taken to be $i$, according to https://en.m.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers. This source also gives a definition for principal square root on all the negative and complex numbers, noting that some principles from the positive numbers no longer hold.