How do I prove that in any ordered field if $x\ne 0$ then $x^2>0$?

Question

I'm asked to prove

In any ordered field if $x\ne 0$ then $x^2>0$?

I thought of doing separate cases for $x>0$ and $x<0$ but I cannot prove that for any $a,b>0$ in an ordered field, $ab>0$ and for $a<0$ in an ordered field $-a>0$. If I'm able to prove this assertion then I think I can prove the main problem.

Any suggestions please?

Answer 1

$x^2$ cannot be $=0$ because there are no zero-divisors. If $x<0$ then $-x>0$ and $(-x)x<0$, that is $-x^2<0$, so $x^2>0$. If $x>0$ then $x^2=x*x>0$.