I have been in a debate over 9gag with this new comic: "The Origins"

"-1 doesn't have a square root?" "Here come imaginary numbers"

And I thought, "haha, that's funny, because I know $i = \sqrt{-1}$".

And then, this comment cast a doubt:

There is no such thing as sqrt(-1). The square root function is only defined for positive numbers. Sorry...

This wasn't by ignorance of complex numbers. In the short round of arguing that happened there, the guys insisted that negative numbers don't have square roots, and that $i$ is defined such that $-1 = i^2$, which is not the same as $i = \sqrt{-1}$. In their opinion, too, no respectable math textbook should ever say that $i = \sqrt{-1}$; which is precisely how Wolfram MathWorld defines $i$.

Is there a consensus over if negative numbers have a square root? Are there some gotchas or disputes with the definitions I should be aware of in the future?

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    $\begingroup$ See the fifth most upvoted question on this site. $\endgroup$ – Alex Becker May 12 '12 at 22:02
  • $\begingroup$ @Alex Becker, there was a general consensus in the thread that complex numbers do exist. The arguing was over the definition of the square root. $\endgroup$ – zneak May 12 '12 at 22:05
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    $\begingroup$ Probably better to say that $-1$ has a pair of square roots, $i$ and $-i.$ That is what happens, it turns out that the square root function must be taken to have two values if extended all the way around the origin in the complex plane. I'm just saying. $\endgroup$ – Will Jagy May 12 '12 at 22:16
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    $\begingroup$ @WillJagy Sorry, not right. The positive numbers also have two square roots. Do not forget that $-2^2 = 4$. The inverse of $x^2$ is not a single-valued function, period. Not in the real domain, not in the complex domain. $\endgroup$ – Kaz May 13 '12 at 0:10
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    $\begingroup$ possible duplicate of What's bad about calling $i$ "the square root of -1"? $\endgroup$ – J. M. is a poor mathematician May 13 '12 at 3:45

I think Asaf's answer, while correct, misses some of the point. It's fairly clear from context that the OP wants to know whether writing $i=\sqrt{-1}$ makes sense in the complex numbers. This is essentially a matter of convention. You can define it that way, but any way you define the function $\sqrt{z}$ it won't have all the properties that it does for real numbers.

A square root of a number $a$ is any number $x$ with $x^2=a$, or equivalently a root of the polynomial $x^2-a$. The fundamental theorem of algebra implies that every complex number $a$ has a square root. In fact, for $a \ne 0$, $a$ has precisely two square roots, which are additive inverses.

You can already see that this is a bit of a problem in the non-negative real numbers. For this case, we choose $\sqrt{a}$ to be the unique non-negative square root, which has a lot of nice properties. Viewed as a function of $a$, $\sqrt{a}$ is continuous, and is a multiplicative homomorphism (i.e. $\sqrt{ab}=\sqrt{a}\sqrt{b}$). These properties are nice enough that it makes sense to call this choice of $\sqrt{a}$ the square root of $a$, instead of a square root of $a$. Of course, this is already abusing terminology a bit, but don't worry too much about that.

It is perfectly reasonable to try to extend the square root function $\sqrt{a}$ for $a$ any complex number, since we know that complex numbers have square roots. But unlike in the positive reals, there's no really nice way to choose what the square root of $a$ should be. In particular, for $\sqrt{-1}$, we can choose either $-i$ or $i$, and since complex conjugation preserves all the algebraic properties of $\mathbb{C}$ we shouldn't expect a nice way to do so purely based on algebraic considerations (like we had for the nonnegative reals).

Let's ignore this for a minute. Any complex number $z$ can be written in the form $z=r e^{i \theta}$ where $0 \le r < \infty$ and $0 \le \theta < 2 \pi$ , the polar representation for complex numbers. For $z \ne 0$ the choice of $r$ and $\theta$ is unique. We can define $\sqrt{z}= \sqrt{r} e^{i \theta /2}$, which is a square root of $z$. Indeed, if we do this, then $\sqrt{-1} = i$. So what's the issue with this?

For one thing, you lose the homomorphism property. There are cases where $\sqrt{ab} \ne \sqrt{a}\sqrt{b}$. This is generic for all extensions, of the square root function, too. There is no we could avoid it by choosing a different definition. You can look at this question to see why this must fail generically.

Furthermore, our choice is not continuous, since $\lim\limits_{y \to 0^+} \sqrt{x-iy} = -\sqrt{x}$ for $x,y$ real. We could make it continuous, but at the cost of not defining $\sqrt{z}$ for $\theta =0$. This is what's called a branch cut. But this is precisely the case when $z$ is a positive real number! Some people do this, but it's bad practice and probably confusing. We could define a branch cut elsewhere, for instance, only allow $-\pi < \theta < \pi$ and use the same formula. Now the branch cut is on the negative real axis, which is better since our notation doesn't conflict with the notation for real numbers, but now $\sqrt{-1}$ is undefined, and $\lim\limits_{y \to 0^+} \sqrt{-1+iy} = i$ while $\lim\limits_{y \to 0^-} \sqrt{-1+iy} = -i$. If you don't care about continuity at the negative real axis, you can extend the definition to $\theta = \pi$, which again gives you back $\sqrt{-1}=i$. We could also induce a branch cut other places, for instance on the positive imaginary axis, to avoid both the problem of disagreeing with the real square root and of being undefined or discontinuous on some part of the real axis, but then your branch cut has to change under complex conjugation, which can also be a problem.

There are other ways to address the issue. The nicest is the theory of Riemann surfaces. The idea here is that you think of $\sqrt{z}$ as a function on a set that is larger than just the complex plane. The Riemannn surface for square root is essentially 2 copies of the complex plane, split at the branch cuts and glued to each other. Here is an image, taken from Wikipedia, for the Riemann surface for the square root function:Riemann surface for the square root

This approach is not without fault either, since you're now no longer talking strictly about the square root of a complex number. The points on the Riemann surface tell you which square root to pick, and there is one point for each square root. Since -1 is a complex number, not a point on the surface, $\sqrt{-1}$ doesn't make sense. However, there are 2 points corresponding to -1, one of which has square root $i$ and the other one $-i$.

You can also consider $\sqrt{z}$ to be a multivalued function, returning a pair of numbers which are both of the square roots of $z$. This works fine, except for when you want to do any sort of actual calculations. In particular, in this approach, $\sqrt{-1} = \{i, -i\}$. This approach does have the homomorphism property that $\sqrt{ab} = \sqrt{a}*\sqrt{b}$, once you define what it means to multiply sets (namely, $\{a,-a\} * \{b,-b\} = \{ab,-ab\}$). This definition does not agree with our definition for positive reals, but it's not as bad as before since the square root of a real number is at least an element of the set of its square roots as a complex number.

EDIT (to clear up an issue in the comments)

I will say, though, that saying that $i = \sqrt{-1}$ is the definition for $i$ is unacceptable in my view. You can't define $i$ this way because it makes no sense. Unfortunately, this is exactly how MathWorld "defines" it, but I think it's circular and doesn't make sense.

$\sqrt{\phantom{-1}}$ is a function which is defined on the non-negative real numbers. $\sqrt{-1}$ does not exist in this context. The whole reason to construct the complex numbers is to fix this, so that you can solve equations like $x^2 = -1$. Before you can say "$i$ is a square root of $-1$" you need to make sure that there is some context in which -1 has a square root. Of course, we all know that the complex numbers are a consistent system with this property, but for hundreds of years people were not sure about this.

But even once you have constructed $\mathbb{C}$, I would argue that defining $i = \sqrt{-1}$ is bad practice. Sure, from an algebraic standpoint, the two square roots of $-1$ are indistinguishable. So there's nothing wrong with defining $i$ to be a square root of $-1$.

The issue here is that $\sqrt{\phantom{-1}}$ is not some universal operator which takes square roots in any context. It's a function on $\mathbb{R}_{\ge 0}$, which we may or may not want to extend to $\mathbb{C}$. Writing $\sqrt{-1}$, without defining $\sqrt{\phantom{-1}}$ on $\mathbb{C}$, is sure to lead to confusion. In fact, whoever wrote the MathWorld article doesn't want to define $\sqrt{z}$ to be just any square root of $z$, for each complex $z$. They at least want continuity on some relatively large subset of $\mathbb{C}$, and some other relatively nice properties. The fact that there are so many ways to define $\sqrt{z}$ for complex $z$ is reason enough to be explicit about what is meant when one writes things like $\sqrt{-1}$.

It's just not clear what $\sqrt{z}$ means until you have defined it. The proper order to do it is this: First, you construct the complex numbers. Second, you choose a square root of -1 and call it $i$ (the other root is obviously $-i$). Finally, after you know what $\mathbb{C}$ and $i$ mean, now you can define what $\sqrt{z}$ means for complex $z$ (if you even want to). At this point, it may or may not be true that $\sqrt{-1}=i$, but if it is this is the definition of $\sqrt{-1}$, not of $i$. I'll also note that it isn't immediately clear that you can even take square roots of arbitrary complex numbers; this is a fact that has to be proven.

I realize this is seems pedantic, and once you're familiar with field extensions and know the existence of algebraic closures for arbitrary fields you can abuse language and define things like $i=\sqrt{-1}$ since you know there is a consistent way to do it (especially if you're not worried about continuity and such). But at the level of someone seeing complex numbers for the first time, it's just too likely to cause some confusion. Of course, people did this sort of thing without explicitly constructing the complex numbers for quite a long time, but no one was really sure that the complex numbers were mathematically consistent. Now that we know they are, they should be presented as such.


tl;dr: There are many ways to generalize the square root function to the complex numbers, some of which have $\sqrt{-1} = i$. However, all of them lack some properties of the real square root function. Different people have different preferences. Some people prefer to reserve the symbol $\sqrt{\phantom{-1}}$ for positive reals, and just talk about a square root of a complex number. For them, the properties of the real square root other than just being a solution to $z^2=a$ are too important to use the same symbol. Some people care less about continuity, being defined everywhere, etc., and they will freely write $\sqrt{-1} = i$. It is a matter of personal preference. You can use whatever convention you like, but need to make sure to be consistent and don't write things like $\sqrt{ab} = \sqrt{a}\sqrt{b}$ when they don't hold, or not write $\sqrt{z}$ where it is undefined. What definitely always holds, regardless of convention, is that $i^2 = -1$.

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    $\begingroup$ +1: you hit on nearly everything I was going to put into an answer, and a bunch more. The only thing I'd add is that while many (but not all) textbooks that define a principal square root function do so using $0 \le \theta < 2 \pi$ (as you talked about first), some books and every calculator and computer algebra system I've ever seen that takes roots of complex numbers defines the principal square root using $-\pi < \theta \le \pi$ (as you talk about second [though I think $\le\pi$, not $<\pi$]). $\endgroup$ – Isaac May 12 '12 at 23:29
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    $\begingroup$ That's a fairly impressive answer! Thank you. $\endgroup$ – zneak May 12 '12 at 23:29
  • $\begingroup$ @Isaac Both are, of course, valid definitions. I wanted continuity, which is why I didn't define it for $\pi$. If you do, you lose continuity, but for computational purposes this is less important than being everywhere defined. The issues are exactly the same as with the $0 \le \theta < \pi$ choice. $\endgroup$ – Logan M May 12 '12 at 23:34
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    $\begingroup$ Yes, I think the principal branch is generally taken as $(-\pi,\pi]$ for the argument, continuity be damned. Addendum: a convenient convention in abstract algebra is $$AB=\{ab:a\in A,b\in B\},$$ in which $\sqrt{ab}=\sqrt{a}\sqrt{b}$ would be universally true, weirdly. $\endgroup$ – anon May 12 '12 at 23:43
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    $\begingroup$ Remembering that old times that I felt boring learning these. Thanks for making them interesting for the first time! most-stupid comment in this question btw. $\endgroup$ – Ayesh K May 13 '12 at 3:49

The question whether something has a root or not must include a setting. Definitions do not appear magically, they require some preexisting framework.

In the setting of the real numbers negative numbers do not have a square root. In the setting of the complex numbers negative numbers do have a square root.

However this is not only when you ask yourself about square roots of negative numbers.

  • Does $2$ have a square root? In the rational numbers it does not. In the real numbers it does.

  • Does $2$ have a multiplicative inverse? In the integers it does not. In the rational numbers it does.

  • Does $2$ have an additive inverse? In the natural numbers it does not. In the integer it does.

These analogies can be stretched into finer details and carried out in other parts of mathematics. The thing to take from that would be that talking about "existence" requires some sort of framework, which is something that the layman is often oblivious to, and accepts the mathematical world as a fixed entity.

But after all, $\frac12$ does not really exist in the physical world, it is an idea - a mathematical idea - and so is $\sqrt{-1}$.

My experience tells me that when someone tells you something like:

There is no such thing as sqrt(-1). The square root function is only defined for positive numbers. Sorry...

It is going to be nearly impossible to convince him otherwise, and it's even less likely to do so on the internet.

  • $\begingroup$ Why the downvote? $\endgroup$ – Asaf Karagila Jan 9 '13 at 10:45
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    $\begingroup$ The problem I have with this answer: there is a distinct difference between "has a square root" and "has the square root". The latter implies a choice; you have to make that choice in the definition of the endomorphism $\sqrt{\cdot}$. The choice can't be made in a good way when your domain extends to negative numbers, because you lose useful properties, such as multiplicative homomorphism and continuity (as explained by Logan Maingi). The cost of making the choice is much too large for what you gain. $\endgroup$ – Rhymoid Jan 9 '13 at 10:49
  • $\begingroup$ Tinctorius, but this question answers something different. The question asked whether or not $\sqrt{-1}$ even exists, or can be defined. You can make a choice in the form of selecting the branch of $\mathrm{Sqrt}$ as a complex function. But to say "only positive numbers have a square root" is devoid of context. Because one can easily say that $2$ has no additive inverse or multiplicative inverse, because in some context this claim is true. $\endgroup$ – Asaf Karagila Jan 9 '13 at 10:54
  • $\begingroup$ That's not how I read the question. It is literally about the consensus about the definition of a square root for negative numbers (it's safe to assume the OP is talking about subsets of $\mathbb{R}$). However, the OP misses that defining the square root function is a different concept. I think addressing this difference should be the core of a good answer. Re-reading the question with that confusion in mind, it is about the consensus regarding the definition of the square root function for negative numbers, and if there are problems with $i = \sqrt{-1}$. You answer neither question. $\endgroup$ – Rhymoid Jan 9 '13 at 11:14
  • $\begingroup$ No. The question asked whether negative numbers have a square root or not. Let me ask you a question, does every polynomial has a root? Oh wait, I didn't even tell you if that's over real numbers, or the complex numbers, or the algebraic closure of $\mathbb F_p$, or over a noncommutative ring of cardinality $\aleph_4$. CONTEXT. It's important. You may say "Oh, but you haven't answered the actual underlying question which the OP had in mind!" but I'm sorry to inform you that I am not a mind reader. I read the question and pointed out the flaw in the arguments: lack of context. $\endgroup$ – Asaf Karagila Jan 9 '13 at 11:17

The square root of a number $k$ may be regarded as the root of a polynomial of the form $z^2 - k$. The fundamental theorem of algebra tells us that we can write $z^2 - k$ as $(z - m)(z - n)$, where $m$ and $n$ are the roots which this function necessarily has, being of degree two. $z$, $k$, $m$ and $n$ are all complex. It's possible that $m$ and $n$ are the same: that is there is a root with multiplicity two. For $z^2 - k$ this happens when $k = 0$.

So basically, for all $k$ everywhere on the complex plane, $z^2 - k$ has two roots, meaning that $k$ has two square roots, except for $k = 0$, where $k$ has two roots that are both equal to zero.

Square root is defined exactly as much for the real numbers as for the complex numbers. There are two square roots of four, namely ${-2, 2}$ and there are two square roots of $-1$, namely ${-i, i}$.

It is irrationally inconsistent to accept that there is a defined square root over the non-negative real number line, but not elsewhere.

(That kind of thinking seems like a remnant of the suspicion that imaginary numbers themselves are some kind of fraud, a view that is outdated by several hundred years.)

  • $\begingroup$ "a remnant of the suspicion that imaginary numbers themselves are some kind of fraud" - the terms "imaginary" and "complex" certainly do not at all help to dispel the notion that these are beasties with untoward behavior. Too bad it seems too late to rename these concepts... $\endgroup$ – J. M. is a poor mathematician May 13 '12 at 4:24
  • $\begingroup$ Yes, I had a flash of the same thought as I wrote that. The somewhat badly named ones are the "irrational" numbers (but of course it makes sense as numbers which are not ratios). All numbers are imaginary in a way, and if a number is actually two, that's a case for "complex". $\endgroup$ – Kaz May 13 '12 at 4:31

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