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The equation "$i^2 = -1$" has no real solution, so there's an entire number system of imaginary numbers that satisfy this.

So why is there not imaginary numbers that satisfy "$|x| = -1$"? There aren't any numbers, real or imaginary, that satisfy that equation, so shouldn't that make a new system of imaginary numbers?

Sebastiano
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    Absolute value is non-negative – J. W. Tanner Jul 30 '21 at 23:25
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    The definition of modulus (absolute value) for complex numbers is equivalent to their "distance" from the origin in the complex plane. Distance is defined not to be a negative number, so there is simply no solution to such an equation. –  Jul 30 '21 at 23:25
  • $|x| = -1$ is not consistent with the requirements of a metric. – econbernardo Jul 30 '21 at 23:27
  • For the same reason why for example $\lfloor x \rfloor = \frac{1}{2}$ has no solutions. $|x|$ is a non-negative real number by definition, while $-1$ is not, so the two can never be equal. If you use a different definition for $|x|$ then you should state that definition in the question. – dxiv Jul 30 '21 at 23:27
  • I think, "imaginary" numbers are real in the sense of existence. They are used in many real life problems. As the solution of $x^2+1=0$ is defined by $i$ (the imaginary unit), you can define the solution of $|x|+1=0$ by $a$ (the "amazing" unit). But the "amazing" unit must have useful applications. – Hussain-Alqatari Jul 30 '21 at 23:32
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    We do not need to, and they are not helpful. You see, complex numbers can be used for various stuff, in maths as well as outside math. A solution to $|x|<0$ won't lead us anywhere. – ultralegend5385 Jul 30 '21 at 23:33
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    $x^2$ also only has nonnegative values. The square function has been extended, though, to include non-positive outputs. The OP's question is more about how this absolute value function cannot also be extended. – David P Jul 30 '21 at 23:44
  • @Hussain-Alqatari but to have an "amazing" unit you must define what $|x|$ means. If $\sqrt{w}$ is number $v$ so that $v^2 = 1$ and we invent $i$ so that $i^2 =-1$ (which isn't actually what mathematicians did) then $|w|$ must means something. And we must invent something where $|w|$ can be negative. But $|w|$ means distance $w$ is from the origin (or words to that effect) and we have to rewrite metrics so that we can have negative distances. – fleablood Jul 30 '21 at 23:48
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    Related (duplicate?): "Is there a number whose absolute value is negative?". Also: "Defining $|x|=-1$". And "Why is there no value for $x$ if $|x|=-1$?". Probably more. – Blue Jul 30 '21 at 23:49
  • Although the old question https://math.stackexchange.com/questions/239794/ was about inventing solutions for differential equations, not for the equation $|x|=-1$, the first paragraph of my answer also applies to the latter equation (and generally to "unsolvable" equations). – Andreas Blass Jul 30 '21 at 23:56
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    I think the big difference is that there is nothing in the definition of the square root that says the input has to be non-negative. But the entire definition of absolute value is based that it is never negative. That the square of all real numbers are non-negative means if we want a negative square root we need a non-real number. So invent one. To have number whose absolute vaule is negative we need a number that when made positive.... is still negative. That's self-contradictory. – fleablood Jul 31 '21 at 00:00
  • $|x|,$ is always a non-negative real number even though $,x,$ is a complex number, hence $,|x|=-1,$ cannot have any solution. – Antonio Jul 31 '21 at 01:18
  • Also related: https://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility – Hans Lundmark Jul 31 '21 at 18:42

2 Answers2

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There are many problems if we define a number to be solution of this equation (some of which are already given in comments above):

  1. Let $x$ be a number such that $|x|=-1$. Then $x^2=1\implies(x-1)(x+1)=0$. Since both the terms on LHS are non-zero, the number system you're defining isn't even an integral domain.

  2. $i^2=-1$ makes sense: since $-1$ shows one step backward on number line ($180^\circ$ rotation), and $i$ shows $90^\circ$ rotation in argand plane (which is essentially an extension of number line in 2D). There is no significance to the one you're suggesting. A point can't have negative distance from the origin.

  3. Again as suggested in comments, it won't be a metric then.

Martund
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$|x|\,$ is always a non-negative real number even though $\,x\,$ is a complex number, hence $\,|x|=-1\,$ cannot have any solution.

So far there does not exist any number which satisfies $\,|x|=-1\,$ and, what is more, if there existed such a number, it would not be an element of a field.

Moreover, if there existed such a number, then the following properties

$1)\quad |ab|=|a||b|\;\;,$

$2)\quad |a+b|\leqslant |a|+|b|\;\;,$

would be no longer valid, indeed if they were correct, we would get that

$|-x|=|(-1)\cdot x|=|-1||x|=1|x|=|x|\;\;,$

$0=|0|=|x+(-x)|\leqslant|x|+|-x|=2|x|=-2\;\;,$

but it is a contradiction.

Antonio
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    This misses the point. $x^2$ was always a nonnegative number until complex numbers were invented. – spaceisdarkgreen Jul 31 '21 at 01:28
  • You are right, but so far there does not exist any number which satisfies $,|x|=-1,$ and, what is more, if there existed such a number, it would not be an element of a field. – Antonio Jul 31 '21 at 01:34
  • Yeah, that's a good reason, although giving up structure isn't necessarily a dealbreaker (e.g. the complex numbers are no longer ordered). – spaceisdarkgreen Jul 31 '21 at 01:46
  • The complex numbers are no longer ordered, but they are still a field. Anyway I do not understand why my answer was downvoted. Are there mistakes? – Antonio Jul 31 '21 at 01:50
  • I didn't downvote it, but it was probably for the same reason I initially criticized it, that it didn't really address the actual question. Editing in the remark that this would ruin the field structure (assuming some reasonable rules for the absolute value) is an improvement though. – spaceisdarkgreen Jul 31 '21 at 02:02
  • @spaceisdarkgreen, I have improved my answer, so I think I do not deserve a downvote anymore. If you agree, you could give me an upvote. – Antonio Jul 31 '21 at 02:25