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Wikipedia mentions the following arithmetic fallacy: $$x<0 \land y <0 \land \sqrt{x \times y}= \sqrt{x} \times \sqrt{y}, $$ since this would lead to $-1=i^2=1$.

So, the above rule is ommitted from the rules of arithmetic of complex numbers.

Now, if we instead of thinking of arithmetic operators as functions, we consider them as relations, then would it be possible to keep the above rule?!

Let's adopt the notation $$a \ * \ b \leadsto c$$ to mean: $c$ is a result of applying operator $*$ on $a, b$.

So, $a \times b \leadsto c$ means: $c$ is a result of $a \times b$.

Then we may maintain the above rule and have: $$ i^2 \leadsto c \iff [c=-1 \lor c=1]$$

Would this lead to an inconsistency in arithmetic of complex numbers?

Adam Rubinson
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Zuhair
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2 Answers2

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What works is to stop worrying and learn to love that the equation $w = \sqrt{z}$, when rewritten as $w^2=z$, defines a relation on $\mathbb C$ whose graph $\{(w,z) \in \mathbb C^2 \mid w^2=z\}$ is a very interesting mathematical object to study.

Lee Mosher
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  • True! But, I think the equation $w=\sqrt z$ when $z \neq 0$ is not correct, it doesn't tell the compelet story; the correct exposition is $\pm w =\sqrt {z}$ or you can put it my way $ (\sqrt {z} )\leadsto w$, since this allows for multivaluation. The problem is with the identity in $w = \sqrt {z}$, this would entail that there is only one number that results from square rooting z, which is not the case. – Zuhair Dec 08 '22 at 18:20
  • $w^2=z$ is not the same as $w = \sqrt {z}$. We don't have $w^2 =z \iff w=\sqrt{z}$. What we have is: $w^2=z \iff (\sqrt{z}=w \lor \sqrt{z}=-w)$, which is usually abbreviated as $w^2=z \iff \sqrt{z}=\pm w$ – Zuhair Dec 08 '22 at 18:41
  • The trouble with your comment is that $\sqrt{z}$ is not well-defined for a complex number $z$. When one wants to do algebra in the complex numbers in a rigorous and unambiguous manner, those abbreviations that many of us use (like for square roots of positive numbers) are usually and wisely avoided when $z$ is a complex number. – Lee Mosher Dec 08 '22 at 19:18
  • Instead, as I wrote, the equation $w = \sqrt{z}$ is usually immediately converted to $w^2=z$ and the unconverted equation $w=\sqrt{z}$ is tossed aside. – Lee Mosher Dec 08 '22 at 19:22
  • Yes, you can stick to $w^2=z$, and informally and loosely thihk of it as $w=\sqrt {z}$, I understand. The correct way of using the "$\sqrt{}$" abbreviation over complex numbers is to adopt the $\leadsto$ notation I've presented here, so $w^2=z$ fully translates formally to "$\sqrt{z} \leadsto w \land \sqrt{z} \leadsto -w$", that's the correct way of doing it formally since here $\sqrt{}$ is a multi-valued RELATION. We can think of this INFORMALLY as $\sqrt{z} = \pm w$. – Zuhair Dec 08 '22 at 21:14
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Multiplication is a clear simple single-valued function. I wouldn't try to make it multi-valued. However, raising complex numbers to non-integer powers is inherently multi-valued, and we can salvage the identity by taking this into account.

If we define $\sqrt{\alpha}$ to be the set of all valid square roots, i.e. $\sqrt{\alpha} := \{ z \in \mathbb{C} : z^2 = \alpha\}$, then the identity works. (You need to also define the product of two sets in the obvious way: $A \cdot B = \{ab : a \in A, b \in B\}$.)

As an example, take $\alpha = -4$ and $\beta = -9$. Then we have $$\begin{align} \sqrt{\alpha \beta} &= \sqrt{36} = \{-6, 6\} \\ \sqrt{\alpha}\sqrt{\beta} &= \{-2, 2\} \cdot \{-3, 3\} = \{-6, 6\} \end{align}$$

The downside of this approach is that it's just annoying to deal with sets of numbers. So understanding the multi-valued nature of $\sqrt{z}$ (or any other $z^\alpha$ with $\alpha \not \in \mathbb{Z}$) is valuable, but when writing out computations people normally just use a single square root branch and accept that this identity is no longer valid.

David Clyde
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    Let $\alpha=-1$ and $ \beta=-1$. Then we have: $$\sqrt{\alpha\beta}=\sqrt{-1 \times -1} = \sqrt {1} = {-1,1} \ \sqrt{\alpha} \sqrt{\beta}= \sqrt{-1} \sqrt{-1} = {-i, i} . {-i,i} = {-1, 1}$$, it works! – Zuhair Dec 07 '22 at 13:24
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    @Zuhair You may also be interested in hyperrings/hyperfields. To be clear, these focus on making addition multivalued so they don't directly address this question, but they run into the same issues of working with multivalued/set-valued functions that this answer mentions ("it's just annoying to deal with sets of numbers"). More generally, this all falls under the umbrella term hyperstructures. – Noah Schweber Dec 17 '22 at 18:59