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Quite a few times on this site, we've been asked about the following calculation: $$ 1 = \sqrt1 = \sqrt{(-1)\cdot (-1)} = \sqrt{(-1)}\cdot \sqrt{(-1)} = i\cdot i = -1 $$ The usual answer is "This is why you should be careful with square roots of things that aren't non-negative reals", or some variation of that.

I have a personal, hand-wavey view of the matter, though, and it goes along these lines: Because we have two square roots of $(-1)$ in the expression, and $(-1)$ has two square roots with no way of distinguishing them, the most natural thing is that one of the square roots becomes $i$ and the other one becomes $-i$. That way, the argument really goes $$ 1 = \sqrt1 = \sqrt{(-1)\cdot (-1)} = \sqrt{(-1)}\cdot \sqrt{(-1)} = i\cdot (-i) = 1 $$ which resolves the paradox (I don't know whether it would be most natural to have the minus sign appear in front of the second square root, or after the evaluation, but I've gone for the latter option here).

This doesn't give the problem $$ 1 = \sqrt1 = \sqrt{1\cdot 1} = \sqrt{1}\cdot \sqrt{1} = 1\cdot (-1) = -1 $$because when the radicand is positive, there is a distinguished root: the positive one.

The most immediate problem, however, is that the square root is supposed to be a function. However, that's somewhat discarded when regarding complex roots at any rate (we're talking about roots of unity, in plural, for instance). Also, we don't choose a root, we pick all roots simultaneously, and the order doesn't matter since complex multiplication commutes.

My question is, is there a sense to this, or is it just wishful thinking, and it would be best if I never mentioned it to anyone ever again?

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    $\begingroup$ "Because we have two square roots of $(-1)$ in the expression, and $(-1)$ has two square roots with no way of distinguishing them, the most natural thing is that one of the square roots becomes $i$ and the other one becomes $-i$" To me, that the same notation would be used to denote two different objects, is most unnatural. $\endgroup$ – Did Jan 12 '17 at 17:10
  • $\begingroup$ @Did You're referring to the fact that $\sqrt{-1}$ would be both $i$ and $-i$ at the same time? I think about it this way: We have two $\sqrt{-1}$ multiplied together; this makes one of them (I don't know which one) into $i$ and the other into $-i$. Or, if you want, it makes the product of the two into $-(i\cdot i)$. At any rate, yes, I'm aware that the use of $\sqrt{\vphantom 1\hphantom 1}$ is unfortunate, but notation is not the main point here. Notion is. $\endgroup$ – Arthur Jan 12 '17 at 17:13
  • $\begingroup$ There are many versions of this question on this site. Here's just one: math.stackexchange.com/questions/990394/… $\endgroup$ – Ethan Bolker Jan 12 '17 at 17:14
  • $\begingroup$ Related: math.stackexchange.com/questions/2080479/… $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 17:15
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    $\begingroup$ What happens if you try to generalise this to $\sqrt i$? Then $i=\sqrt i\sqrt i "=" \frac{1+i}{\sqrt 2}(-\frac{1+i}{\sqrt2}) = -i$. I don't think there is a way of making your notion well defined. $\endgroup$ – Mathmo123 Jan 12 '17 at 17:15
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If you'd like you can see the square root as a multi-valued function $\sqrt{\cdot} : \mathbb{C} \to \mathbb{C}$. In other words it's an actual function $\mathbb{C} \to \mathcal{P}(\mathbb{C})$, where $\mathcal{P}$ denotes the power set.

Now if $R$ is a ring, then you can define an addition on $\mathcal{P}(R)$ by $$A+B = \{ c \in R \mid \exists a \in A, b \in B \text{ s.t. } c = a + b \}$$ and similarly a multiplication by: $$A \cdot B = \{ c \in R \mid \exists a \in A, b \in B \text{ s.t. } c = a \cdot b \}.$$

The unit for the addition is $\{0\}$, and the unit for the multiplication is $\{1\}$. This is not quite a ring though, since there are no inverses for $+$ (e.g. $\varnothing$ has no inverse, nor does any set with at least two element): it's a (commutative) semiring. There is an embedding of semirings $j : R \to \mathcal{P}(R)$ given by $j(z) = \{z\}$.

Now you can define the square root $\sqrt{A}$ to be $$\sqrt{A} = \{ b \in \mathbb{C} \mid \exists a \in A \text{ s.t. } b^2 = a \}.$$ It's now clear that for any nonzero $z \in \mathbb{C}$, $\sqrt{j(z)}$ is a two-element set, containing the two complex square roots of $z$. For example $\sqrt{j(1)} = \{+1, -1\}$ (and $\sqrt{j(0)} = j(0)$). And now when you compute $(\sqrt{j(-1)})^2$, you get $$\bigl(\sqrt{j(-1)}\bigr)^2 = \sqrt{j(-1)} \cdot \sqrt{j(-1)} = \{i,-i\} \cdot \{i,-i\} = \{1,-1\}$$ and $1$ belongs to that set. But it's really abusive to say that it is equal to $1$, since in some sense $-1$ is another "solution". And now you're dealing with multivalued functions, semirings that aren't rings and all that nice stuff, where all algebraic intuition breaks down (for example "$0 \cdot x = 0$" is not true: $\{0\} \cdot \varnothing = \varnothing$!).

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  • $\begingroup$ A minor quibble: starting at "For example", the notations are not quite consistent since every $\sqrt{z}$ in the post should actually read $\sqrt{\{z\}}$ (or $\sqrt{j(z)}$). Two options: either stick to the construction based on the embedding $j$ and replace $\sqrt{z}$ by $\sqrt{\{z\}}$ (or $\sqrt{j(z)}$) everywhere, that is, five times, I believe; or explain once and for all that $\sqrt{z}$ stands for $\sqrt{\{z\}}$, and then keep in mind that $\sqrt{4}$ is $\{2,-2\}$, not $2$). Which option is best from a pedagogical standpoint is debatable. (+1 anyway.) $\endgroup$ – Did Jan 13 '17 at 7:40

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