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According to the definition of absolute value negative values are forbidden.

But what if I tried to solve a equation and the final result came like this: $|x|=-1$ One can say there is no value for $x$, or this result is forbidden.

That reminds me that same thinking in the past when mathematical scientists did not accept the square root of $-1$ saying that it is forbidden.

Now the question is :"is it possible for the community of math to accept this term like they accept imaginary number.

For example, they may give it a sign like $j$ and call it unreal absolute number then a complex number can be expanded like this:

$x = 5 +3i+2j$ , where $j$ is unreal absolute number $|x|=-1$

An other example, if $|x| = -5$, then $x=5j$

The above examples are just simple thinking of how complex number may expanded

You may ask me what is the use of this strange new term? or what are the benefits of that?

I am sure this question has been raised before in the past when mathematical scientists decided to accept $\sqrt{-1}$ as imaginary number. After that they knew the importance of imaginary number.

SchrodingersCat
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Pentapolis
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  • A major difference is that squareroots can be defined more simply: The squareroot of $-1$ is some thing which squares to $-1$. The nice thing is that taking squares is something we under stand much better than squareroots. In your example there does not seem to be any similar "opposite" operation. – Tobias Kildetoft Jul 07 '16 at 12:44
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    Absolute value is also defined as the distance between the point and the origin (for both complex and real numbers). Considering that while mathematicians work with non-euclidean geometry they introduced a sphere with negative and imaginary radii, your question actually makes sense. – Levent Jul 07 '16 at 12:45
  • Would the absolute value deserve the name "absolute value" if it could give a negative output? In my personal opinion, not really. In your opinion? I don't know. – Arthur Jul 07 '16 at 12:58
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    The reason people started using imaginary numbers was that they led to solutions of problems - involving real numbers - that they were unable to solve without taking square roots of negatives. Furthermore, the solutions arrived at via imaginary numbers actually worked in $\mathbb{R}$. Do you know of any problem in the familiar number system that can be sensibly solved using numbers with negative absolute value? If so, then I predict your new numbers will catch on. :) – G Tony Jacobs Jul 07 '16 at 13:02
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    Related: http://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility – Hans Lundmark Jul 07 '16 at 13:07
  • An other example $\ln|jx|=\pi i+\ln|x|, x \in \Bbb R$ – Pentapolis Jul 07 '16 at 14:13

2 Answers2

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The beauty of math is that you can define everything. The question is: what properties you want this "j" to satisfy? For example, I guess that you want the absolute value $|\cdot|$ to satisfy the triangle inequality. Note that $$ 0=|0|=|j+(-j)|\leq|j|+|-j|=-1-1=-2 $$ a contradiction.

boaz
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  • Why would $j\in\mathbb{C}$ be a contradiction? The whole point would be to "expand" the reals to get something like this. – Tobias Kildetoft Jul 07 '16 at 12:53
  • @TobiasKildetoft $j\in\Bbb C$ contradicts $z\in\Bbb C\implies |z|\ge0$. – Mario Carneiro Jul 07 '16 at 12:56
  • I'm not sure how you get that $|j^2|=1$ implies $j^2\in\Bbb C$, though (which is implied by your $a+bi$ decomposition). – Mario Carneiro Jul 07 '16 at 12:57
  • The absolute value for complex number is by definition a non negative number. So if you want to define a "new number" as "j", you need to expand the number system. In order to define $|j|=-1$, you need to give a complete different definition for absolute value. – boaz Jul 07 '16 at 12:57
  • Actually, never mind my previous comment. Where did you get that it belonged to $\mathbb{C}$? There does not actually seem to be any argument for that. – Tobias Kildetoft Jul 07 '16 at 12:57
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    A possibly bigger problem is that $a+bi+cj$ decomposition implies that this new space is a $3$-dimensional algebra over the reals, which can't be a field. – Mario Carneiro Jul 07 '16 at 13:00
  • Your edit, showing how we get the contradiction $0\leq -2$ shows us that, if we are going to allow negative values for absolute value, then the triangle inequality has to go. – G Tony Jacobs Jul 08 '16 at 00:24
  • If the triangle inequality relies on a geometric meaning, how can you prove it works when triangles are no longer constructable? What about if triangles did indeed exist somehow for such cases in a different way, then we'd have a new triangle inequality that matches the new values and simplifies down to the old one whenever we aren't using $j$? – Simply Beautiful Art Oct 02 '16 at 12:34
  • The triangle inequality is the last thing one would want to keep. For instance, it does not hold and makes no sense in split-complex numbers and tessarines. – Anixx Sep 24 '21 at 17:58
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Absolute value is a special case of the definition of norm. We need to impose some properties on it so that the domain of the function (or the space which the norm is defined on) can have special structures. The properties of a certain function result from a specific purpose.

Xianjin Yang
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