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We know that how a single definition $i^2=-1$ revolutionized our mathematics and solved many many problems. I wonder whether the definition $|p|=-1$ could have the potential of creating a new generation of numbers and help us in other areas like complex numbers do(from geometry to calculus).

Has anyone researched on it?

Please add appropriate tags.

Zev Chonoles
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evil999man
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    This question is somewhat related: http://math.stackexchange.com/questions/259584/why-dont-we-define-imaginary-numbers-for-every-impossibility/259605#259605 – Alex Petzke Apr 30 '14 at 12:57

2 Answers2

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There is a difference here.

The value $i$ is defined as the number that solves the equation $x^2+1=0$. The reason that this equation does not have a solution in $\mathbb R$ is that for every $x\in\mathbb R$, $x^2>0,$ which is a consequence of the properties of the real numbers. There is nothing inherit in the equation that would demand it to have no solution.

On the other hand, the value $|x|$ is defined to always be positive, thus it will by definition never equal to $-1$.

The difference then:

  • $x^2>0$ is a consequence of the properties of real numbers. Looking at numbers with different properties may change this fact.
  • $|x|>0$ is inherit in the definition of $|\cdot|$. It is a property that must hold for all numbers, even if we expand our set.
Irregular User
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5xum
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  • What if there could be a better definition of || which we now just know as a special case? – evil999man Apr 30 '14 at 09:25
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    That is of course possible, but I do not know of any way in which that would be useful. – 5xum Apr 30 '14 at 09:43
  • So based on the second sentence in your answer, I would like to ask you a similar question then: Can we define a number $j$ that solves the equation $\frac{1}{x}=0$? In fact, I will post it as a question myself... – barak manos Apr 30 '14 at 09:54
  • @barakmanos That would be a lonely number I guess. Btw was it sarcasm? – evil999man Apr 30 '14 at 10:00
  • @Awesome: No sarcasm here. What do you mean "lonely"? If we define $j$ as the solution to the equation $\frac{1}{x}=0$, then we can use $\frac{1}{2}j$ as the solution to the equation $\frac{2}{x}=0$, etc. – barak manos Apr 30 '14 at 10:03
  • I sense that both $j$ will be same. Might be a mistake. – evil999man Apr 30 '14 at 10:05
  • @5xum That was people must have said when defining complex numbers. How the hell will this help us. It is just some pure mathematical concept with no application. – evil999man Apr 30 '14 at 10:13
  • @Awesome Indeed. That is exactly why I used the same phrazing. I dear not say that it will not be useful. I find it entirely plausible that it may, at some point, be useful. I am merely saying that it is not the properties of numbers that cause the trouble (as was the case with complex numbers) but properties of the absolute value itself. – 5xum Apr 30 '14 at 10:20
  • But 4|z|^2 = ((z+conjugate of z)^2+(iz-i*conjugate of z)^2) – Taemyr Apr 30 '14 at 10:29
  • @Taemyr true, but any function that will expand the absolute value will not be a norm any more. It will have to skip at least one of the axioms for norms. – 5xum Apr 30 '14 at 11:07
  • Just a comment (not really a response to anyting)

    There are many definitions of |z|-like functions. See metric spaces(http://en.wikipedia.org/wiki/Metric_space). None of them allow -ve though.

     – Frames Catherine White Apr 30 '14 at 11:21
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    I think it's misleading to nonsensical to say that the positiveness of $|\cdot|$ "must hold" no matter how we extend it. Absolute value is simply not defined on any other set, so by extending it we have to just make up a definition. There's absolutely nothing stopping you from defining it to be negative on some other set. I don't think that the distinction you're describing is real, it's just a subjective difference. – Jack M Apr 30 '14 at 11:24
  • @5xum However your answer makes no mention of norms and as such does not indicate that the salient fact is that || is not a norm. Also, if I remember correctly, the concept of norms was developed as a consequence of complex analysis, which came about from the study of i. – Taemyr Apr 30 '14 at 11:25
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    @JackM That is more or less exactly 5xum's point. – Taemyr Apr 30 '14 at 11:26
  • If we think of it as the distance from the origin, then we can ask: can we invent a space that has negative distance? Like maybe extending a new dimension from every point in space to have a space of uncountably infinite dimension. I don't know if that is possible, but maybe someone else has some immediate insight. – jdods May 14 '16 at 15:50
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You can indeed modify the absolute value in $\Bbb Q$ to construct completions of it which are structurally very different from $\Bbb R$. This is how to define the $p$-adic numbers which depend on a preliminary choice of a prime number $p$. Nonetheless, even these "exotic" absolute values are always non-negative valued.

Andrea Mori
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