$\sqrt{-1}$ is both a positive and a negative number

Question

I contend that there is a third category of number (in addition to positive and negative numbers), which are neutral.

For the sake of expression, let us call these numbers neutral numbers. Zero, for example, is a neutral number.

The reason I believe this is as a mathematical extension of philosophical dialetheism: the belief that that some statements can be true and untrue simultaneously (such as a: (a) is false).

The square root of $-1$ is $i$. $i$ squared is $-1$.

$i$ is an "imaginary" number because we cannot place it on our number line. No number on the number line squared gives $-1$. Nevertheless, we know that (humour me) the value of $i$ is 1, since $1/1$ or $1\cdot 1$ or $-1\cdot 1$ etc. always gives plus or minus 1.

The only number with "value 1" which when squared gives $-1$ would be a number which is neither positive nor negative, in such a way that $i^2 = 1\cdot(-1)$, or rather $(\pm 1)\cdot (\pm 1)$.

I propose that where the $x$-axis shows positive and negative number beside $0$, there is also a $z$-axis that shows neutral numbers in such a way that $i = \pm 1$.

I know that this appears as a crazy idea, but I came to it though philosophy not mathematics. Despite how crazy it is, is this possible mathematically?

Thank you.

Answer 1

So I take it we all agree on $i=\sqrt{-1}$. From there, whether or not this is crazy hinges on whether there is any point to having $i=\sqrt{-1}$, which arises from casus irriducibilis.

That is, one cannot factor the polynomial $x^3-3x-1$ without these numbers, even though the roots do indeed exist on the number line. Thus, there is a relationship between these and the real numbers.

Answer 2

There is no ordering in the complex domain. I will show this by contradiction.

Lets assume $i>0$, then if we multiply by $i$, which is positive (hence the inequality sign does not change).

$i^2>0\cdot i=0$, but $i^2=-1$, hence we get a contradiction.

Now, assume $i<0$. Again, multiply by $i$, which is now negative (hence the inequaltiy sign will change).

$i^2>0$. Again, we get a contradiction. Hence, we cannot say if a complex number is positive or negative.