Defining $|x|=-1$

Question

Q 1a

Is it possible to define a number $x$ such that $|x|=-1$, where $|\cdot|$ means absolute value, in the same manner that we define $i^2=-1$?

I have no idea if it makes sense, but then again, $\sqrt{-1}$ used to not be a thing either.

To be more explicit, I want as many properties to hold as possible, e.g. $|a|\times|b|=|a\times b|$ and $|a|=|-a|$, as some properties that seem to hold for all different types of numbers (or in some analogous way).

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Q 1b

If we let the solution to $|x|=-1$ be $x=z_1$, and we allow the multiplicativeness property,

$$|(z_1)^2|=1$$

Or, further,

$$|(z_1)^{2n}|=1\tag{$n\in\mathbb N$}$$

Note that this does not mean $z_1$ is any such real, complex, or any other type of number. We used to think $|x|=1$ had two solutions, $x=1,-1$, but now we can give it the solution $x=e^{i\theta}$ for $\theta\in[0,2\pi)$. Adding in the solution $(z_1)^{2n}$ is no problem as far as I can see.

However, there result in some problems I simply cannot quite see so clearly, for example,

$$|z_1+3|=?$$

There exists no such way to define such values at the moment.

Similarly, let $z_2$ be the number that satisfies the following:

$$|z_2|=z_1$$

As far as I see it, it is not possible to create $z_2$, given $z_1$ and $z_0\in\mathbb C$.

The following has a solution, in case you were wondering.

$$|\sqrt{z_1}|=i$$

so no, I did not forget to consider such cases.

But, more generally, I wish to define the following numbers in a recursive sort of way.

$$|z_{n+1}|=z_n$$

since, as far as I can tell, $z_{n+1}$ is not representable using $z_k$ for $k\le n$. In this way, the nature of $z_n$ goes on forever, unlike $i$, which has the solution $\sqrt i=\frac1{\sqrt2}(1+i)$.

So, my second question is to ask if anyone can discern some properties about $z_n$, defining them as we did above? And what is $|z_1+3|=?$

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Q 2a

This part is important, so I truly want you guys (and girls) to consider this:

Can you construct a problem such that $|x|=-1$ will be required in a step as you solve the problem, but such that the final solution is a real/complex/anything already well known. This is similar to Casus irreducibilis, which basically forced $i$ to exist by establishing its need to exist.

I am willing to give a large rep bounty for anyone able to create such a scenario/problem.

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Q 2b

And if it is truly impossible, why? Why is it not possible to define some 'thing' the solution to the problem, keep a basic set of properties of the absolute value, and carry on? What's so different between $|x|=-1$ and $x^2=-1$, for example?

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Thoughts to consider:

Now, Lucian has pointed out that there are plenty of things we do not yet understand, like $z_i\in\mathbb R^a$ for $a\in\mathbb Q_+^\star\setminus\mathbb N$. There may very well exist such a number, but in a field we fail to understand so far.

Similarly, the triangle inequality clearly cannot coexist with such numbers as it is. For the triangle inequality to exist, someone has to figure out how to make triangles with non-positive/real lengths.

As for the properties/axioms of the norm I want:

$$p(v)=0\implies v=0$$

$$p(av)=|a|p(v)$$

Answer 1

First of all, you can define $|\cdot|$ to mean whatever you want in any given context, as long as you're clear and upfront about it.

That being said, one usually wants $|\cdot|$ to be a norm, which means it fulfills a certain list of criteria. Among them is $|x|\geq 0$. If you break these rules, does your operation really deserve to be called "absolute value"? Does your operation deserve to be written using $|\cdot |$? Personally, I would say it doesn't, which means that using that symbol wouldn't be wrong, per se, but it would make it more difficult for your readers to understand what's going on, simply because of what they expect from that notation.

One notable exception, as pointed out in the comments, is the determinant of square matrices. And real / complex numbers are square matrices (of dimension $1\times1$), so in that context we really have $|-1|=-1$. But that's a different operation.

Answer 2

The absolute value is quite a different thing than a square. A square simply comes from multiplication and nothing else. Especially, a square does not need an order on the underlying structure. However, the absolute value can only be defined after an order in defined by setting $$ |x| = \begin{cases} x & x\geq 0\\ -x & x < 0\end{cases}. $$ So, it is indeed defined to be non-negative. It is not that you may have some algebraic structure with an absolute value and then ask yourself "What if $|x|$ is negative?" in the same way you ask about squares… Put differently:

You can't deduce form the field axioms that $x^2 = -1$ has no solutions, but you can deduce from the axioms of the ordering that $|x|=-1$ has no solutions.

To answer the actual question: I haven't seen variant of absolute values (or norms, or metrics) to take negative values and doubt that such a thing has been studied.

Answer 3

What about extending the real number absolute value to complex numbers in a different way?

For $z = re^{iθ}$, let $|z| = re^{2iθ}$

You still get that the "new" absolute value for all reals matches the regular one.

With this, we have $|i| = 1\cdot e^{2i(π/2)} = e^{iπ} = -1$.

Answer 4

When doing operations on complex or split-complex numbers, one can encounter a result which has negative modulus. In that case we consider it equal to a number with positive modulus but argument shifted by $2\pi$. It is also of note that in split-complex numbers (tessarines) the modulus can be imaginary: $|a+bj|=\sqrt{a^2-b^2}$. Thus, $|j|=i$.