Showing that $i$ is neither negative nor positive

Question

Show that $i$ is neither negative nor positive.

Proof:

Assume that $i<0$

$i×i>0×i$ Since $i<0$

$i^{2}>0$

Since $i^{2}:=-1$

So, $-1>0$ which doesn't hold and hence this is a contradiction to the fact.

Next, assume that $i>0$

$i×i>0×i$ since $i>0$

$i^{2}>0$

Since $i^{2}:=-1$

So, $-1>0$ which again doesn't hold and hence this is a contradiction to the fact.

This establishes the proof.

Is this valid proof?

Is there any other way to prove this? Also pls check my proof. Thanks. — Waseem F-Name Ali Nawaz, May 26 '21 at 13:16
Does this answer your question? Positive and negative complex numbers? — , May 26 '21 at 13:17
This is the standard way to show that there is no order in the complex numbers. You established it correctly. In particular, neither $i>0$ nor $i<0$ makes sense, hence $i$ is neither positive nor negative. — Peter, May 26 '21 at 13:19
What I mean to ask that is i means iota neither negative nor positive? So I have proved it. I wanna know your suggestions. — Waseem F-Name Ali Nawaz, May 26 '21 at 13:20

score 1 · Accepted Answer · answered May 26 '21 at 13:23

Your proof is correct. More generally one might say:

Assume $k$ is an ordered field. Then every square must be non-negative, since $$x<0\implies x^2>0,x>0\implies x^2>0$$ for every $x\in k$. Thus $\mathbb{C}$ cannot be an ordered field, since it is algebraically closed and thus all elements are squares.

Showing that $i$ is neither negative nor positive

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