# $\sqrt{-1}$ is both a positive and a negative number [closed]

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.

• For the sake of everyone's sanity, square rooting usually refers to the positive solution of $x^2=a\qquad$ – Simply Beautiful Art Nov 6 '16 at 18:00
• I think you may have reinvented the Argand plane... – Parcly Taxel Nov 6 '16 at 18:00
• This is less a philosopy question and more a terminology question. Rather than try to force the terminologies "positive number" and "negative number" to apply beyond the context of the real number line, mathematicians have (for a very long time) followed the simpler method of not using those terminologies in the context of the complex plane. – Lee Mosher Nov 6 '16 at 18:10
• It depends on the $z^{1/2}$ branch cut you choose ( in a particular case ). – Felix Marin Nov 8 '16 at 4:23

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.

• But I mean this in such a way that i = 0 in the standard plane. Just another kind of 0. Do you understand what I mean? – Bonj Nov 6 '16 at 18:18
• $i=0$ makes no sense because $i^2=0^2=0\neq -1$. – MrYouMath Nov 6 '16 at 18:19
• Okay yeah. i cannot be zero, but it also cannot be greater or smaller than zero. i must be different in a different way, which perhaps we have not yet a mathematics to explain? Have I lost my mind? – Bonj Nov 6 '16 at 18:25
• Normally, in mathematics we give up the conpect of greater or smaller if we deal with complex numbers. We can only say that the magnitude of a complex number $|z|$is positive or zero. – MrYouMath Nov 6 '16 at 18:27

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.