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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).


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 $|z_1+3|=?$


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.


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?


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:

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

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

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  • $\begingroup$ Is it possible for the length of an object to be negative? $\endgroup$ – Math1000 Oct 2 '16 at 1:22
  • $\begingroup$ If anyone has some good tags for this question, that'd be nice. $\endgroup$ – Simply Beautiful Art Oct 2 '16 at 1:22
  • $\begingroup$ hi, i asked the same a while ago: math.stackexchange.com/questions/1844930/… $\endgroup$ – SAJW Oct 2 '16 at 1:23
  • $\begingroup$ @Math1000 I don't know, depends on the object. $\endgroup$ – Simply Beautiful Art Oct 2 '16 at 1:23
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    $\begingroup$ For what it's worth, $|.|$ is sometimes used to denote the determinant of a matrix, and there certainly are matrices with negative determinant. $\endgroup$ – Joey Zou Oct 2 '16 at 1:31
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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.

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  • $\begingroup$ I guess, I do not understand: why cannot we just define it as $k$ (like we did $i$) or some equivalent and carry on? $\endgroup$ – heather Oct 2 '16 at 1:46
  • $\begingroup$ @heather You mean "let's invent a new number $k$ such that $|k|=-1$, and see what happens"? You could do that. I don't know what happens. What would $|k+1|$ be, for instance? $\endgroup$ – Arthur Oct 2 '16 at 8:00
  • $\begingroup$ $0$, maybe? I don't know. Are you saying that doesn't work? $\endgroup$ – heather Oct 2 '16 at 11:27
  • $\begingroup$ @heather No, I'm saying I don't know what happens. $\endgroup$ – Arthur Oct 2 '16 at 11:39
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    $\begingroup$ @heather I don't think so, otherwise it probably would've been done already, and well known. $\endgroup$ – Arthur Oct 2 '16 at 11:55
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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.

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  • $\begingroup$ $-x$ is not actually negative here. And such definitions often times do not make sense, like if $x\in\mathbb C$, for example. $\endgroup$ – Simply Beautiful Art Oct 2 '16 at 1:35
  • $\begingroup$ @SimpleArt Yes, what you are saying reflects that there is not ordering of the complex numbers… So if you ever want absolute values to be negative, you need to find a new meaning of "absolute value" and probably one not based on an ordering. $\endgroup$ – Dirk Oct 2 '16 at 1:38
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    $\begingroup$ Well, to think back for a moment, how did we define $\sqrt{-1}$? We never really gave it any new meaning, we just said "Hey, I'll call this number '$i$' and use all the same properties as before!" $\endgroup$ – Simply Beautiful Art Oct 2 '16 at 1:43
  • $\begingroup$ @SimpleArt you are a mathimatical gangster hehe ;) $\endgroup$ – SAJW Oct 2 '16 at 1:45
  • $\begingroup$ @SimpleArt ok, i deleted it, just saying i'm with you on defining new meanings to something and other's did too before you, as you pointed out yourself. $\endgroup$ – SAJW Oct 2 '16 at 1:48
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The other answers talk about reals or norms, but do not consider complex numbers. However, if you think of $|x| = (x^2)^{\frac{1}{2}}$, then $|x|=-1$ does not have a solution either. I therefore do not think there is a natural extension to the negatives.

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  • $\begingroup$ Perhaps you mean $\vert x \vert = \left({x \bar{x}}\right)^\frac 1 2$ $\endgroup$ – GFauxPas Oct 2 '16 at 12:26

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