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What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!

So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?

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    $\begingroup$ Déjà vu... $\endgroup$ – J. M. isn't a mathematician Dec 10 '10 at 13:30
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    $\begingroup$ This is just as in the real case when we choose $\sqrt{4}=2$ instead of $\sqrt{4}=-2$. $\endgroup$ – Joe Johnson 126 Dec 10 '10 at 15:24
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    $\begingroup$ @Joe : in the real case there are informal arguments that a square root of 2 should exist. This is the opposite of the situation with -1, where prior concepts suggest that a square root should NOT exist. It took thousands of years to get to complex numbers after having some notion of square root. $\endgroup$ – T.. Dec 11 '10 at 5:34
  • $\begingroup$ @Alexei Averchenko : It is not plainly false, as you said since it is arbitrary, one very well can opt for the other root, and everything would still hold true by simply multiplying by -1 (or rotating the complex plane by 180 degrees ). Having two things being true at the same time does not lead to a contradiction. $\endgroup$ – jimjim Dec 11 '10 at 12:28
  • $\begingroup$ @T.. : Is that a typo ? did you meant "formal arguments" rather than "informal" ? Should the temporal order in which humans discovered the complex number system should be considered important? One could very well start with $\mathbb{C}$ and consider $\mathbb{R}$ within $\mathbb{C}$ instead. $\endgroup$ – jimjim Dec 11 '10 at 12:37
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I believe this is a common misconception. Here in Sweden we (or at least I) was taught that $i^2 = -1$, not that $\sqrt{-1} = i$. They are two fundamentally different statements. Of course most of the time one chooses $\sqrt{-1} = i$ as the principal branch of the square root.

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    $\begingroup$ @Arjang The two statements $i^2 = -1$ and $\sqrt{-1} = i$ are not equivalent statements. Suppose they were equivalent. Then one would also have to say that $(-i)^2 = -1$ and $\sqrt{-1} = -i$ are equivalent statements. Since both $i^2 = -1$ and $(-i)^2 = -1$ are true, their (supposedly) equivalent statements must also be true, i.e. $\sqrt{-1} = i$ and $\sqrt{-1} = -i$. But this implies $i = -i$, which is a contraditction. Therefore the supposed equivalence is false. $\endgroup$ – Alex Basson Dec 11 '10 at 14:48
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    $\begingroup$ @Arjang: We've gone through this before: $\sqrt{}$ is a function, not short-hand for "is a number whose square is equal to". $\sqrt{4}=-2$ is incorrect if you are using the symbol to denote the function, while "$-2$ is a number whose square is $4$" is correct. We do not "know for sure" that $\sqrt{4}=-2$ and $\sqrt{4}=2$ is correct; in fact, you are advocating that we should eliminate transitivity of the "=" symbol, a far more radical departure than the simple convention that $\sqrt{}$ represents a function. $\endgroup$ – Arturo Magidin Dec 11 '10 at 19:55
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    $\begingroup$ @Arjang: You wrote: "We know for sure that '$\sqrt{4}=2$ and $\sqrt{4}=-2$' is true." [emphasis added]. So you are either claiming that $2=-2$, or that = is not transitive. As to where is it stated that $\sqrt{}$ is a function, in the same place where it is stated that $2$ is $S(S(0))$ and that $4$ is $S(S(S(S(0))))$, where $S$ is Peano's successor function that $0$ is the primitive. In any case, you seem to want to conflate $\sqrt{n}$ with "solution to $x^2=n$." The two are not the same (yes, by (very widespread) convention; if you want to avoid it, the onus is on you to say so). $\endgroup$ – Arturo Magidin Dec 11 '10 at 21:44
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    $\begingroup$ @Arjang: If the conjunction is true, then either = is not transitive, or $2=-2$; under the standard, widespread convention, $\sqrt{4}=-2$ is not true, because it is not the principal value of the square root function. You continue to conflate the square root function with "solutions to $x^2=4$." If you want to not abide by the standard convention, that's your right, but you have to explicitly say so. When you say "$\sqrt{4}=2$ AND $\sqrt{4}=-2$" is obviously true, then you are saying either that $2$ and $-2$ are equal, or that = is not transitive. There is simply no way out of that. $\endgroup$ – Arturo Magidin Dec 11 '10 at 22:58
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    $\begingroup$ @Arjang: There are many ways in which you can make the statement that there are two solutions to $x^2=4$ without either implying somewthing obviously problematic, or violating standard conventions. You can talk about the different branches of the (complex) square root function, for instance. If you simply say "$\sqrt{4}\in\{2,-2\}$", then most people will not understand what you are trying to say; instead, they will say "Well, yes; it is also true that $\sqrt{4}\in\{2,1\}$; so what?" What you want is to talk about solutions to the equation (which is fine!) not about $\sqrt{}$. $\endgroup$ – Arturo Magidin Dec 12 '10 at 3:35
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$\sqrt{-1}$ is an imprecise notation. There are several ways of making it precise, some of which involve what comes down to an arbitrary choice of square root of $-1$ and some of which don't. This is because there are several related ways to construct $\mathbb{C}$ from $\mathbb{R}$:

  • As the ring $F = \mathbb{R}[x]/(x^2 + 1)$ (edit: following in the discussion in the comments, perhaps it would be better to say "an algebraic closure of $\mathbb{R}$."). The Galois group $\text{Gal}(F/\mathbb{R})$ has order $2$, and complex conjugation $x \mapsto -x$ is its only nontrivial element. The Galois symmetry here forbids us from distinguishing between $x$ and $-x$.

  • As the ring $\mathbb{C} = \mathbb{R}[\imath]/(\imath^2 + 1)$, where we fix the choice $\imath$ of generator. This is what is typically meant by $\mathbb{C}$. This field has a distinguished element $\imath$ satisfying $\imath^2 = -1$, and this is what we usually mean by $i$. $\mathbb{C}$ is isomorphic to $F$, but not canonically so, since $\imath$ can be sent to either $x$ or $-x$ and there is no way to choose between these.

  • As the ring $R$ of linear endomorphisms $\mathbb{R}^2 \to \mathbb{R}^2$ of non-negative determinant preserving an inner product $\langle \cdot, \cdot \rangle$. There are two elements of $R$ squaring to $-1$, corresponding to a rotation and its inverse, and there is no way to choose between them unless $\mathbb{R}^2$ is also equipped with an orientation, which is an identification of its exterior square with $\mathbb{R}$. This choice of orientation corresponds to the difference between the first and the second constructions above.

But most people will not bother to stop and talk about such subtleties as $F, \mathbb{C}, R$ being non-canonically isomorphic, so they say $\sqrt{-1} = i$ because the truth is (for practical purposes) unnecessarily complicated.

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    $\begingroup$ In Number Theory Seminar, Hendrik Lenstra put it this way: "Suppose the Martians defined the complex numbers by adjoining a root of $-1$ which they called $j$; and when the Earth and Martians start talking, the scientists have to translate $i$ to be either $j$ or $-j$; we'll decide it's $j$, because that's probably what the scientists will decide..." About 10 minutes later in the seminar, he said, "But then it was discovered that most Martians are left-handed, so the philosophers decide that's it better to send $i$ to $-j$ instead. Well, no matter." $\endgroup$ – Arturo Magidin Dec 11 '10 at 2:37
  • $\begingroup$ @Arturo: Reading that and the stuff on your website, it seems Lenstra is every bit as awesome as you say he is. $\endgroup$ – J. M. isn't a mathematician Dec 11 '10 at 4:30
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    $\begingroup$ I don’t follow your assertion that $F$ and $C$ are not canonically isomorphic. They’re not uniquely isomorphic to each other over $R$ — just like each has non-trivial Galois group, i.e. is not uniquely isomorphic to itself over $R$. But there is a distinguished choice of isomorphism, by their construction: they have canonical generators $x$, $\iota$. Indeed, as far as I can see, your construction of the two is (up to variable-naming) exactly the same; I don’t see what distinction you’re intending? $\endgroup$ – Peter LeFanu Lumsdaine Dec 11 '10 at 6:00
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    $\begingroup$ @Peter: sorry if I was unclear. In the first construction I am not distinguishing the generator x. I didn't know a good way to indicate this. Perhaps I should say simply say "F is an algebraic closure of R." $\endgroup$ – Qiaochu Yuan Dec 11 '10 at 6:10
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    $\begingroup$ OK, I guess I was being a little intentionally obtuse; I do see the distinction you’re intending to make, but I don’t see mathematically how your original phrasing makes it. “$F$ is an algebraic closure of $R$” is a very very different approach, but I think a nice one — by black-boxing the construction it really does force us to forget any distinguished generator. $\endgroup$ – Peter LeFanu Lumsdaine Dec 11 '10 at 6:17
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As even the capitalistic Wikipedia acknowledges, there are three laws of dialectics:

  1. The law of the unity and conflict of opposites;
  2. The law of the passage of quantitative changes into qualitative changes;
  3. The law of the negation of the negation.

Thus, from law 1, we see that i and -i are both equal and opposite. From law 2, we learn that $\mathbb C$ is qualitatively different from $\mathbb R$. From law 3, we attain the dynamic equilibrium between i and -i that was so powerfully expounded by the revolutionary hero Évariste Galois.

This at least is how we taught such matters in the USSR.

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    $\begingroup$ The most hilarious answer I've seen on this site, thanks. :-) $\endgroup$ – ShreevatsaR Dec 11 '10 at 6:35
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    $\begingroup$ In Soviet Union contradiction proves you! $\endgroup$ – jimjim Dec 11 '10 at 12:46
  • $\begingroup$ Editing this answer seems to be a yearly ritual for you. :) $\endgroup$ – user856 Aug 25 '12 at 7:08
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    $\begingroup$ @Rahul, that reflects my pace of learning -- I just figured out how to link to a paragraph :-) I don't want anybody thinking I made those laws up, do I? $\endgroup$ – TonyK Aug 25 '12 at 8:01
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This is answered quite succinctly in the MathWorld article on Principal Square Root.

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  • $\begingroup$ In that article would it made more sense that the final stament to be : there is no ambiguity in defining as "the principal" square root of -1 ,rather than : there is no ambiguity in defining as "the" square root of -1? $\endgroup$ – jimjim Dec 11 '10 at 12:10
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    $\begingroup$ @Arjang, If we add the word principal then we should also take away the scare quotes. It is generally understood that when we speak of "the square root of 4" that we mean principal square root, as opposed to "a square root of 4" which includes both square roots. It is a common use or abuse of language that is tolerated because it is easy enough to understand what people mean by it. $\endgroup$ – Mitch Schwartz Dec 11 '10 at 20:36
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The proper way to define $i$ is to define the laws for complex addition and multiplication the usual way using ordered pairs of real numbers, and then define $i$ to be $(0,1)$. This distinguishes it from $-i$ which is $(0,-1)$.

One can then set the root of $-1$ to be $i$, but of course this last step is pretty arbitrary. You can see it either as wrong or as not wrong if you want to work with the last step as an arbitrary definition. The important thing is that we have an unambiguous definition of $i$ using the initial steps specified above.

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  • $\begingroup$ There is no proper way of defining $i$ there are many proper ways of defining $i$. It is nothing more than convention that i is (0,1) or (1,0). One system is obtained by rotating the other 180 degrees. Everything else stays the same. $\endgroup$ – jimjim Dec 11 '10 at 13:13
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Alex it is just a notion. We use the value of $i = \sqrt{-1}$ so that we can work more freely. For example consider the Quadratic Equation $$x^{2}+x+1=0$$ The discriminant for this is $D=b^{2}-4ac=-3$. So the roots have the value $$x = \frac{-1 \pm{\sqrt{3}i}}{2}$$ which looks better when written with an $i$ notation. That's all.

I don't know how this is taught in the US but to me, i encountered this when i was at high school, learning how to solve for Quadratic Equations when the Discriminant is less than $0$.

Next, note that $\mathbb{C}$ doesn't have the same ordering as $\mathbb{R}$. That is for any 2 real numbers $a,b$ we have either:

  • $a>b$

  • $a< b$

  • $a=b$

But for complex number's this is not true. Since if you take $i$ and $0$, we must have either $i > 0$ or $i < 0$, but this isn't true.

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    $\begingroup$ @Chandru1: You should either get rid of the minus sign, or of the $i$ in your second displayed equation. $\endgroup$ – Arturo Magidin Dec 10 '10 at 17:22
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    $\begingroup$ @Chandru1: One can easily give an order for $\mathbb{C}$. The reason we don't usually do so is because what $\mathbb{C}$ lacks is an order that is compatible with the operations; that is, an order such that for all $a$, $b$, and $c$, if $a\lt b$ then $a+c\lt b+c$; and if $c\gt 0$ then $a\lt b$ implies $ac\lt bc$. $\endgroup$ – Arturo Magidin Dec 10 '10 at 17:24
  • $\begingroup$ we can also write $x^{2}+x+1=0 \Rightarrow (x+\omega)\times(x-\omega^2) = 0$ $\endgroup$ – Quixotic Dec 10 '10 at 21:01
  • $\begingroup$ Using the $i$ or $-i$ does not make the work more freely or otherwise. The only thing that changes is that everything has rotated 180 degrees! nothing is gained or lost at all. I don't understand the statement about looks better written with $i$ notation, better than what? Also what does ordering has to do with taking going clock wise or anticlockwise by convention? $\endgroup$ – jimjim Dec 11 '10 at 12:52
  • $\begingroup$ Actually, changing $i$ to $-i$ in your example changes precisely nothing, and it's not what I was asking about anyway. $\endgroup$ – Alexei Averchenko Dec 11 '10 at 13:31
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This seems to be very little different than the standard notation for $\sqrt{4} = 2$ as opposed to $\sqrt{4} = -2$. By convention, we have a concept of a principal branch of the square root. More importantly, though, is that there is no real difference between letting $i := \sqrt{-1}$ and $-i := \sqrt{-1}$ (though there is an imaginary difference!). One really needs only to notate the two roots of $\sqrt{-1}$ by $\pm i$.

In addition, I first learned about imaginary numbers in a US school and we learned that i was s.t. $i^2 = -1$ as well. As far as I can recall, the first important distinction comes when we consider how to represent the complex plane graphically and through the exponential series - but again, these are merely notational inconveniences rather than big problems.

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Does $\pm i=\pm\sqrt{-1}$ sound any better ? Also, $(-i)^2=-1$, so I don't see how defining it as $i^2=-1$ would make things any better.

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