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?

Answer 1

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.