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I was wondering why if we have $i^2 = -1$, why not have a "number" $v$ such that $|v| = -1$? Does anything interesting arise from considering this system?

The only thing I could come up with was:

$$ |a+bv| \le |a| + |bv| = a - b $$

Kviii
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3 Answers3

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Such a number (or vector) $v$ simply can't exist if you want $|{-}|$ to behave like $|{-}|$ should behave. For example we'd have

$$|1+v| \le |1|+|v| = 1-1 = 0$$ and so $$1 = |1| = |(1+v)-v| \le |1+v|+|(-v)| = |1+v|+|v| \le 0 + (-1) = -1$$ But $1 \le -1$ is nonsense.

In general there is no reason why arbitrarily postulating the existence of a 'thing' satisfying a certain property should yield anything worthwhile. The story behind $i$ is somewhat different.

Clive Newstead
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  • If we define $v$ with these features, then your second line will not be true, will it? You did the similiar thing with $i=\sqrt{-1}=\sqrt{\dfrac{1}{-1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}=\dfrac1i=-i$. –  Dec 19 '14 at 22:34
  • Not so. I used two axioms in the definition of a norm: triangle inequality and homogeneity. Any reasonable meaning of $|{-}|$ would have it be a norm. This isn't the same as with the square root, which is defined algebraically: in that case, the result $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is a result that has to be proved from the hypothesis that $a$ and $b$ are non-negative reals. (It fails in that case because $i$ is not a non-negative real.) – Clive Newstead Dec 19 '14 at 22:39
  • Yes, it is true for all complex numbers and vectors, but why you assumed that $v$ is a complex number? I know that $v$ with this features do not exist, but if we define it, then $v$ cannot be a complex number, it will belong to some new number set. –  Dec 19 '14 at 23:03
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First off, it wouldn't be very interesting to have a norm that was negative for all nonzero vectors in a vector space -- that would just be the same old norm we already know, with a minus in front of it.

So suppose we have vectors $v, w$ with $|v|<0$ and $|w|>0$. Let's further assume that $v$ and $w$ are linearly independent. If they are not, assuming that our vector space is at least two dimensional, let $q$ be some vector that is linearly independent of $v$ (and this also $w$). If $|q|>0$ then use $q$ as the new $w$; otherwise use $q$ as the new $v$.

Now consider the function $$ f(t) = |(1-t)v + tw| $$ This is certainly something that looks like it ought to be continuous on $[0,1]$, but $f(0)=|v|<0$ and $f(1)=|w|>0$, so by the intermediate value theorem, there should be some $t$ such that $|(1-t)v+tw|=0$. And because $v$ and $w$ are linearly independent, $(1-t)v+tw\ne 0$.

Now we have a nozero vector with zero norm, and that wreaks all kinds of havoc for most of the things we want to use norms for.

Bonus chatter: There are in fact spaces with norm-like things $\|{\cdot}\|$ such that $\|v\|^2$ is always real, but can be either positive or negative(!). A prominent example is Minkowski space, which forms the "spacetime" of Special Relativity. Here we do indeed have nonzero vectors with zero $\|v\|$, which play a special role in the theory -- but for that very reason they don't satisfy many of the properties the usual length of a vector has.

hmakholm left over Monica
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The absolute value of a real or complex number is necessarily a non-negative real number (this is a property of norms), and so breaking this property means most of the usual statements involving $|\cdot|$ and $v$ will probably be false.

If you wanted to, however, there's no reason you couldn't take some abstract set $E$, containing (at least) your new "number" $v$, and investigate some function $$\phi : E \cup \Bbb C \to \Bbb R$$ such that $\phi(z) = |z|$ for $z \in \Bbb C$ and $\phi(v) = -1$ for some $v \in E$ (and then you have to decide what else $\phi$ does or what properties you want it to have).

BaronVT
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