Is $i$ equal to $\sqrt{-1}$?

Question

Some of my math teachers said that $i=\sqrt{-1}$ is the wrong definition for $i$ and that the correct definition is $i^2=-1$. If the second definition if true, then does it mean $\sqrt{-1}= ±i$? Which of the two definitions is true?

$i^2=-1$ is not defining anything. Since, if you have a number that satisfies it its opposite also satisfies it. — Hellen, Jul 22 '17 at 13:49
$i=\sqrt{-1}$ is actually more proper. You are picking precisely one new element required to satisfy that $(\sqrt{-1})^2=-1$. — Hellen, Jul 22 '17 at 13:51
See also https://math.stackexchange.com/questions/887724/refining-my-knowledge-of-the-imaginary-number — Hans Lundmark, Jul 23 '17 at 06:37

seldon · Accepted Answer · 2017-07-22T14:10:02.730

7

From a formal algebraic perspective, $\Bbb{C}$ is the field $\Bbb{R}[x]/(x^2+1)$ which is the field you get by adjoining to $\Bbb{R}$ the roots of $x^2 + 1$ and all the linear combinations of the form $a+ib$, where $a, b \in \Bbb{R}$. You really need to add just one of them (we call it $i$), because the other is just $0 + i(-1)$. Notice that it doesn't matter which we pick as $i$: if I pick one and you picked the other, we would never notice we made different choices.

Infact, the Galois group (automorphisms preserving the original field, thus fixing $\Bbb{R}$ in this case) of $\Bbb{C}$ has exactly two automorphisms: the identity, sending $i$ to itself, and the conjugation, sending $i$ to its opposite. By definition, automorphisms preserve structure, thus confirming the choice of $i$ doesn't really matter.

edited Jul 22 '17 at 14:10

answered Jul 22 '17 at 14:02

seldon

1,382

Id est, from your reasoning follows that $\sqrt{-1}={i,-i}$. Am I right? – Michael Rozenberg Jul 22 '17 at 14:09
If by $\sqrt{a}$ we mean the roots of the equation $x^2 - a = 0$, yes. Yet it doesn't make sense to define $i$ that way because as far as $\Bbb{R}$ is concerned, that equation doesn't have roots. – seldon Jul 22 '17 at 14:12
Surely, usually we aren't that pedantic and such a definition meets the right intuition, but if we want to answer questions such as this one, we have to. – seldon Jul 22 '17 at 14:14
But by your way we get something not good. For $\sqrt{2i}={1+i,-1-i}$ you don't remember any another definition and for $\sqrt{-1}$ you make it. What do you think? – Michael Rozenberg Jul 22 '17 at 14:15
Sorry I'm not understanding your point. Are you saying we are defining it for $i$ and using it for $2i$, so that the same symbol means two different things? – seldon Jul 22 '17 at 14:17
I say about $\sqrt{-1}$ only. What is it? – Michael Rozenberg Jul 22 '17 at 14:18
Let us continue this discussion in chat. – seldon Jul 22 '17 at 14:23
Yet, it has been a fresh answer after years! – Mikasa Sep 19 '23 at 07:02

score 4 · Answer 2 · answered Jul 22 '17 at 13:44

4

The square root symbol does not make sense when applied to a negative number. Granted, people often write it that way for convenience. $i$ is defined by the equation $i^2=-1$. Yes, if there is a field containing a root of $x^2+1$ then if $i$ is such a root so is $-i$. When we define a field such as $\mathbb C$ we are implicitly choosing a root of $x^2+1$.

answered Jul 22 '17 at 13:44

lulu

70,402

2

The $\sqrt.$ here is different from the normal $\sqrt. :\Bbb R^{\ge0}\rightarrow\Bbb R^{\ge0}$. We're defining a 'new version' with the same symbol for convenience as it extends the old one. – Shuri2060 Jul 22 '17 at 14:05
@Shuri2060 You can't define $i$ and $\sqrt x$ in the one line. That is, you can't define $i$ in terms of $\sqrt x$ and simultaneously define $\sqrt x$ in terms of $i$. – lulu Jul 22 '17 at 17:45
It depends, but I think $\sqrt.$ is usually first extended so that its domain is $\Bbb R$ with the requirements that it remains a bijection. Then $i$ is a 'shorthand' for $\sqrt{-1}$ much like $2$ is a shorthand for $S(1)$. – Shuri2060 Jul 22 '17 at 17:48
@Shuri2060 Well, I'd define that extension by saying that $x<0\implies \sqrt x=\sqrt {|x| } , i$. With $i$ defined as one of the roots of $x^2+1$ over $\mathbb R$. How would you define it? Is there some nice axiomatic way to do it? Anyway, I have to imagine that the OP means to start with $i$ and then extend the definition of $\sqrt x$, no? – lulu Jul 22 '17 at 17:53
I see - in which case I think your way of defining it would probably be best. But perhaps there is some nice axiomatic way of doing it - I'll think about it. – Shuri2060 Jul 22 '17 at 18:12

Masacroso · Answer 3 · 2017-07-22T13:50:21.457

$\sqrt{\cdot}$ is seen in most cases as a function in $[0,\infty)$, so it have a unique value associated to each non-negative real. In this tradition we generally define $i:=\sqrt{-1}$, that can be thought informally as a kind of extension to the classical square root function.

Indeed we can extend the definition of the square root to any complex number, setting it as the principal value of $\sqrt z:=e^{1/2\ln z}$ with the same result.

However is true that $(-i)^2=-1$.

score 3 · Answer 4 · answered Jul 22 '17 at 15:51

3

$\sqrt{-1}$ is denoted as $i$ and $i$ is a solution of the algebraic equation $x^2+1=0$. That is why $i^2=-1$.

One could define $i=-\sqrt{-1}$ and using this notation he could built the Complex analysis. That is up to him.

answered Jul 22 '17 at 15:51

MAN-MADE

5,381

Is $i$ equal to $\sqrt{-1}$?

4 Answers4