# Could we invent a new number with $|p|=-1$?

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?

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This question is somewhat related: math.stackexchange.com/questions/259584/… – Alex Petzke Apr 30 '14 at 12:57

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.
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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 9:25
That is of course possible, but I do not know of any way in which that would be useful. – 5xum Apr 30 '14 at 9:43
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 9:54
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
@JackM That is more or less exactly 5xum's point. – Taemyr Apr 30 '14 at 11:26

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.

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