Is it true that “there is no such thing as the square root of minus one”?

Is the statement "there is no such thing as the square root of minus one" a true statement?

It seems to me that we need to be careful about the word "the" as it appears in the statement. If we see it as implying uniqueness, then surely the statement is true after all, since $i$ and $-i$ are distinct square roots of minus one.

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I think your question is better suited for the English language and usage stackexchange site. – amWhy Jun 25 '14 at 18:14
It is sort of true, for the reason you point out. However, it is possible to have a convention that "the" square root of the complex number $(-1,0)$ (I am using the ordered pair representation) is $(0,1)$, and not $(0,-1)$. There is after all a similar situation in the reals, where by convention $\sqrt{1}=1$. – André Nicolas Jun 25 '14 at 18:16
@amWhy Don't mathematicians argue over/discuss semantics frequently? An annoying level of precision, even in constructing sentences such as OP's, is usually desirable ... – Zubin Mukerjee Jun 25 '14 at 18:37
@Andre Nicolas Our mathematical conventions are so economical. Perhaps we subconsciously apply Occam's (or, Ockham's) Razor in making these decisions. – Chris Leary Jun 25 '14 at 19:05
It’d be more accurate to use the indefinite article ‘a’; hence, according to my point of view (which I certainly won’t impose on anyone else), $i$ is a square root of $-1$ and $- i$ is a square root of $-1$. If the domain that you’re working with is the set of all positive real numbers only, then you could adopt the convention that the positive square root is the canonical one. – Berrick Caleb Fillmore Jun 25 '14 at 19:38

Indeed, as suggested in the question, if we were to speak of the root of $-1$, then there should be just one such value, or at least one of the two possible values should be distinguished.

For positive reals, we can speak of, for instance, the square root of $4$, because out of the two possible values of $+2$ and $-2$, only one is positive (and this is the one we take). Note that numbers $+2$ and $-2$ are quite different: for instance, $+2$ is a square of a real number, and $-2$ is not.

For $-1$, things are a little trickier. There are two possible choices of the root, $+i$ or $-i$. And in some sense, these choices are both equally legitimate. There is no way to distinguish $+i$ from $-i$ using only arithmetic operations and knowing what the real numbers are. This is a consequence of the fact that the conjugation mat $z \mapsto \bar{z}$ preserves multiplication, addition, and leaves $\mathbb{R}$ invariant.

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Usually we speak of 'the' square root of a negative number $-k$ ($k\in\mathbb R_{>0}$) as $\sqrt{-k}=\sqrt{k}\cdot i$, but we could have chosen $-\sqrt k\cdot i$ just as well. This follows from the fact that the Galois Group of $\mathbb C$ contains two elements (two automorphisms of $\mathbb C$), $e$ and $s$: $$e:\mathbb C\to \mathbb C: c\mapsto c\\ s:\mathbb C\to \mathbb C: c\mapsto \overline {c}$$ This means that $\mathbb C$ would be exactly the same as (undistinguishable from) $s(\mathbb C)$, where $i$ and $-i$ are interchanged.

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This is a good answer, but I suspect that it may be lost on the OP (at least I learned complex before Galois theory, but of course every undergraduate sequence is unique). – Chris K Jun 25 '14 at 18:34
@ChrisK, I know that is probably the case. I just finished a Galois theory course myself, so I just had to write it down. – Ragnar Jun 25 '14 at 18:35

Presumably you do not have an issue with the existence of $i$ and $-i$, and are just debating the grammar of the sentence. Note that the square root function is, in general, defined to be the positive root or, in such cases that sign does not make sense, we use the principal solution. That is, the value of $e^{\pi/2+k\pi}$ with smallest positive argument (namely, $i$).

Using this convention, there is such a thing as the square root of $-1$. But as you note, there are $2$ numbers which square to $-1$.

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It depends on how you define "squareroot"...

But consider rotations around the $z$-axis over an angle $\phi$.

We may denote such rotations as $R(\phi)$, and we have the property $R^2(\phi) = R(2\phi)$.

Now consider $R(180^\circ) = R(-180^\circ)$ - that would be $-1$, so $\textbf{i} = R^2(\pm 90^\circ)$ has the property that $\textbf{i}^2 = -1$.

Or $\sqrt{R(\pm 180^\circ)} = R(\pm 90^\circ)$.

Or $\sqrt{-1} = \pm \textbf{i}$.

Note that the squareroot is double-values as

$$x^2 = z \Rightarrow x = \pm \sqrt{z}$$

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You don't need to write $180^o$. You can write $180^\circ$. I edited accordingly. – Michael Hardy Jun 25 '14 at 18:30
Thanks! I did not knew to use \circ that way!!! – johannesvalks Jun 25 '14 at 18:35
You're welcome. One point of English grammar: Nobody says "I did not knew". One says "I did not know". – Michael Hardy Jun 25 '14 at 18:36
"To use \circ that way, I knew not." – Zubin Mukerjee Jun 25 '14 at 18:38
Well I am not native English - so I make that kind of language errors, sorry for that... - But THANKS for the correction!!!! – johannesvalks Jun 25 '14 at 18:54