If $i^2=-1$ why isn't $i$ also equal to $-\sqrt{-1}$?

Question

I've read on this site several times that it is better to say $i^2=-1$ as opposed to $i=\sqrt{-1}$.

If we let $i^2=-1$, why doesn't $i=-\sqrt{-1}$? This makes sense since $(-\sqrt{-1})(-\sqrt{-1})=-1$.

Does the property $x^2=4\Rightarrow x=\pm2$ not hold here? If so, could someone please explain?

Note; This question is unique because I'm asking just about the fact that since $i^2=-1$, why isn't $i$ equal to $\pm\sqrt{-1}$. User64786 and Dietrich Burde's comment's reveal that what I am asking about is different than the tagged question. It seems that the answer is a matter of convention, not algebra.

Thank you.

We define $i$ to be $+\sqrt{-1}$ and then $-i=-\sqrt{-1}$. Indeed, also $(-i)^2=-1$. — Dietrich Burde, Apr 20 '19 at 15:53
Algebraically there is no difference between $i$ and $-i$. You can call $i$ to be either one of the two roots. It is only an arbitrary geometric convention to choose $i$ to be such that $1,i$ is a frame oriented counterclockwise in the plane and call $-i$ the other root of the equation. — user647486, Apr 20 '19 at 16:00

score 4 · Accepted Answer · 2019-04-20T16:26:34.057

Neither $\sqrt{-1}$ nor $-\sqrt{-1}$ make sense in the reals, and in fact $i=\sqrt{-1}$ is quasi meaningless.

It is conventional to write

$$i^2=-1$$ where we understand $i$ to be an "operator that rotates by a quarter turn", so that when applied twice it has the effect of a half-turn.

Following this logic, we also have

$$(-i)^2=-1.$$

Actually, it is impossible to tell $i$ from $-i$ (the direction of rotation is immaterial), and we stick to the symbol $i$.

score 2 · Answer 2 · answered Apr 20 '19 at 15:54

2

What you're confusing here is variables with constants. $x^2=-1$ has two roots, $i$ and $-i$. $i^2=-1$ holds because $i$ is defined as $\sqrt -1$. $i$ is a constant.

answered Apr 20 '19 at 15:54

Vizag

3,257

score 2 · Answer 3 · answered Apr 20 '19 at 18:53

I guess when people introduced the $\sqrt[n]{x}$ function they worked with real numbers only. As long as we need only real numbers this is a nice function: it's value is a number such that if multiplied by itself n times... you know.

There is a slight problem with this definition. There might be several numbers which satisfy this requirement. And this fact makes it very inconvenient to use this function. Mathematitians decided to modify the definition of root function slightly: "it'a a non-negative number...". This approach seems to be more convenient: if we have $\sqrt{4}$ always equals $2$. And if in some cases we need to account all the possibilities, we will explicitely write $\pm\sqrt{4}$.

Again, it's just very convenient if function has no more than one value and it was very easy to define the root function in such a way.

But things changed when people started to use complex numbers. For any complex number $c$ there are always two numbers $x$ such that $x*x=c$. And there is no way to define a function $\sqrt{c}$ unambiguously. There is no useful rule how to choose one of the two candidates to be the value of "square root" function.

That's why it's better to avoid using $\sqrt[n]{c}$ function with complex numbers unless it is clear which one of the $n$ candidates is supposed to be used.

There are two complex numbers $x$ such that $x*x=-1$. Funny story - these two numbers are pretty much identical (all their properties are absolutely the same!), but they are still different numbers. Mathematitians called one of them $i$. Which one? It's impossible to tell, impossible to describe somehow which one of them was chosen. Just one of them. The other one became $-i$.

I mentioned earlier the general rule to avoid $\sqrt{c}$ then complex numbers are involved, but my personal experience is that in case $\sqrt{-1}$ the second part of the rule applies: it is very clear that $i$, not $-i$ is implied.

This is false 'And there is no way to define a function $\sqrt{c}$ unambiguously. There is no useful rule how to choose one of the two candidates to be the vale of "square root"'. — user647486, Apr 20 '19 at 19:45
One of the many choices that show that that claim is false is this. — user647486, Apr 20 '19 at 20:46
@user647486 Thanks. Indeed this is a consistent way to define a root function. But you mention other choices are possible. So this brings back the problem: it is not clear what is the value of $\sqrt{c}$ in each particular case. That's why it's better to avoid this notation unless it's obvious which value is implied. — lesnik, Apr 20 '19 at 21:33

score 1 · Answer 4 · answered Apr 20 '19 at 16:03

1

We define $i$ as the value $\sqrt{-1}$ and $-i$ as $-\sqrt{-1}$. We have that $x^2+1=(x+i)(x-i)$

In general, we have that $\sqrt{a}=|a^{\frac12}|$ for $a\in \Bbb R$, and $\Im(\sqrt z)>0$ for $z\in\Bbb C$

answered Apr 20 '19 at 16:03

Rhys Hughes

12,842

If $i^2=-1$ why isn't $i$ also equal to $-\sqrt{-1}$?

4 Answers4