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This question already has an answer here:

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.

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marked as duplicate by Mikhail Katz, Watson, C. Falcon, Chill2Macht, Henrik Jul 7 '16 at 16:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – Tobias Kildetoft Jul 7 '16 at 12:44
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    $\begingroup$ 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. $\endgroup$ – Levent Jul 7 '16 at 12:45
  • $\begingroup$ 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. $\endgroup$ – Arthur Jul 7 '16 at 12:58
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    $\begingroup$ 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. :) $\endgroup$ – G Tony Jacobs Jul 7 '16 at 13:02
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    $\begingroup$ Related: math.stackexchange.com/questions/259584/… $\endgroup$ – Hans Lundmark Jul 7 '16 at 13:07
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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.

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  • $\begingroup$ Why would $j\in\mathbb{C}$ be a contradiction? The whole point would be to "expand" the reals to get something like this. $\endgroup$ – Tobias Kildetoft Jul 7 '16 at 12:53
  • $\begingroup$ @TobiasKildetoft $j\in\Bbb C$ contradicts $z\in\Bbb C\implies |z|\ge0$. $\endgroup$ – Mario Carneiro Jul 7 '16 at 12:56
  • $\begingroup$ 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). $\endgroup$ – Mario Carneiro Jul 7 '16 at 12:57
  • $\begingroup$ 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. $\endgroup$ – boaz Jul 7 '16 at 12:57
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    $\begingroup$ 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. $\endgroup$ – Mario Carneiro Jul 7 '16 at 13:00
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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.

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Absolute value for hypercomplex numbers is defined by: $+\sqrt{\sum_i c_i^2}$ where $c_i$ is the ith coefficient. Since the coefficient is a real number, the square of it is positive, sum of positive numbers will be positive. Absolute value takes the positive square root.

Another way to define absolute value is the 'distance' from 0 the point defined by the number is, where the number of coefficients is the number of dimensions.(1,$c_1$,$c_2$... are mutually orthogonal). Distance(towards 0) can only be positive, which should be obvious.

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  • $\begingroup$ Thanks, Ariana. I love "Dangerous Woman", btw. $\endgroup$ – The Count Jul 7 '16 at 13:37
  • $\begingroup$ @TheCount Yay;) Love you $\endgroup$ – Ariana Jul 7 '16 at 13:37

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