Prove using the axioms that $x>0$ implies $-x<0$

Question

How to prove equations that if $x>0$, then $-x<0$ using the axioms of the real numbers $\Bbb{R}$ (if $x \in \Bbb{R}$)?

My university lecturer gave this as an exercise and I am stuck on which axioms to use. I was thinking you could use O4 but that uses the condition that $z>0$ which means you can't set $z=-1$.

Here are the axioms

Please help!!

Answer 1

By Axiom O3, we have that, since $0<x,$ $0+(-x)<x+(-x).$ Now, by Axioms F3, F4, we have that $-x<0.$

Answer 2

If $x, -x>0$, then $0<x = x+0<x+(-x) = x-x=0$. If $-x=0$, then $0=x-x=x+(-x)=x+0=x>0$. In both cases, $0<0$, so we must have $-x<0$.

Answer 3

Use O3: start with $x>0$ and add $-x$ to both sides