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I recognise that a trivial proof exists via contradiction and complex numbers, namely: $$1=\sqrt{-1*-1}=\sqrt{-1}\sqrt{-1}=i^2=-1 $$ Which is obviously untrue therefore the rule does not hold for negative arguments of the radical function. I've been looking around for a direct proof that uses an alternate method, but I have been unable to find such a thing. Specifically, I'm asking if there is a proof that produces $\sqrt{mn} \neq \sqrt{m} \sqrt{n}$ for $m,n < 0$ without contradiction.

The reason I am interested is because if the only way to conclude the above domain for the radical identity is through contradiction, what is to say that the converse isn't true, and that the existence of complex numbers implies such 'logical flaws', and the only way to get around that is to just 'unrigorously cheat' and put a domain on the law.

In a sense I could wrap my head around that: if you think about $n^2$, because a complex number is a solution to $n^2=-1$, in a very crude way, the solutions must be of the formation of $a\times b$ where say, $b\in \Re^{-}$, but somehow maintain equality as it numbers being squared ergo $-1=1$.

Theo Bendit
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Mannix Showell
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    How do you define $\sqrt m$ for negative $m$? If you are thinking of the complex square root: That is multi-valued, and you have to specify a branch, otherwise the question makes no sense. – Martin R Oct 01 '21 at 09:01
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    Would it be ok to write the proof of the example as follows? "Note that $\sqrt{-1 \times -1} = \sqrt{1} = 1$, and $\sqrt{-1} \times \sqrt{-1} = i^2 = -1$. As $1 \neq -1$, we have $\sqrt{-1 \times -1} \neq \sqrt{-1} \times \sqrt{-1}$." – Theo Bendit Oct 01 '21 at 09:01
  • Martin, I define $\sqrt{m}$ for negative $m$ as arbitrarily the positive branch to the solution to $m^{2} = -1$. Hope this helps :)

    Theo, that still leaves a somewhat sour taste because it is the same proof just stated in reverse :/

     – Mannix Showell Oct 01 '21 at 09:28
  • @MannixShowell To show that a rule does not work, a single counterexample is enough. The proof that there is no order in the complex numbers is also given by assuming $i>0$ and $i<0$ and refute both possibilities. Here choosing the other branch does not help since $(-i)^2=-1$ leads to the same contradiction. We could only repair this example by choosing $i$ for one of the roots and $-i$ for the other, that does not really make sense. – Peter Oct 01 '21 at 09:36
  • Right. However, and I say this entirely to be semantic here, what if this could be the fault of assuming that $\sqrt{-1}$ exists? I get that a single counterexample is proof enough, but that surmises that complex numbers should exist -- could this not be proof that complex numbers "while exist, imply logical fallacies"?

    Also I get the other half of your comment :)

     – Mannix Showell Oct 01 '21 at 09:43
  • Theo's proof isn't a proof by contradiction. It directly shows that there exist a pair of numbers ($1$ and $-1$) that don't satisfy the identity. It's still a proof by example, but I'm not sure what other type of proof you're looking for to disprove "for all real m and n, the identity holds". The negation is "there exists some pair m and n such that the identity does not hold", and any direct proof must either be like Theo's or some more arcane argument that allows you to produce examples like the one under discussion. Even the answer by user2661923 is like this, just with a class of examples. – Mark S. Oct 01 '21 at 12:24
  • "But that surmises that complex numbers should exist" I mean, yes, if you decide not to define $\sqrt{-1}$ as $i$ then you can just say "this isn't even defined for all real $m,n$." But not extending an identity isn't a logical fallacy by any stretch. For example, the reals don't extend the property "is a quotient of integers" that rationals have. – Mark S. Oct 01 '21 at 12:26
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    Thanks Mark, I just realized the mistake in understanding I made: a brain-fart, so to speak. I understand now, cheers :) – Mannix Showell Oct 01 '21 at 12:52

1 Answers1

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Expanding on the comment of Martin R, the question is meaningless unless a convention is accepted for the definition of $\sqrt{z} ~: z \in \Bbb{C}, z \neq 0.$

A common Complex Analysis convention, which dovetails with the corresponding Real Analysis convention, is that if $z \neq 0$ and
$z = re^{i\theta} = r[\cos(\theta) + i\sin(\theta)] ~: ~-\pi < \theta \leq \pi$,
then $\sqrt{z} = \sqrt{r}e^{i\theta/2}.$

Here the principal Argument of $z = \theta$, and the principal Argument of $\sqrt{z} = \theta/2.$

Under this convention, you will always have that $-\pi/2 < \text{the principal Argument of} ~\sqrt{z} \leq \pi/2.$

Now consider $z_1 = r_1e^{i\theta_1}, z_2 = r_2e^{i\theta_2} ~:$
$0 < r_1, r_2, ~-\pi < \theta_1, \theta_2 \leq \pi.$

Then $z_1 \times z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}$.

Then $\sqrt{z_1 z_2} = \sqrt{r_1 r_2} \times e^{i\alpha}$,
where $-\pi/2 < \alpha \leq \pi/2$ and $\alpha \equiv (\theta_1/2 + \theta_2/2) \pmod{\pi}.$

Contrast this with $\sqrt{z_1} \times \sqrt{z_2} = \sqrt{r_1 r_2} \times e^{i/2(\theta_1 + \theta_2)}$.

So, $\sqrt{z_1 z_2} = \sqrt{z_1} \times \sqrt{z_2} \iff (\theta_1/2 + \theta_2/2) = \alpha$.

Stating the assertion a different way, if $\theta_1, \theta_2$ are the principal Arguments of $z_1, z_2$,
then $\sqrt{z_1z_2} = \sqrt{z_1} \times \sqrt{z_2} \iff -\pi/2 < (1/2)(\theta_1 + \theta_2) \leq \pi/2.$

user2661923
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