given that $F$ is an ordered field, and $a \in F$ , prove that $a^2 \geq 0$,
Now From the axiom that given $x,y \in F$ either $x \leq y$ or $y \leq x$, its obvious that there is 2 choices for $a$.
either $a \geq 0$ and in this case we know that $a^2 = a*a \geq 0$ because for any $x,y \in F$ if $x \geq 0$ and $y \geq 0$ then $x y \geq 0$.
Help me prove the case when $a \leq 0$ i know that $-a \geq 0$ so $-a *-a \geq 0$ and $(-1)*a = -a = a*(-1)$ , so what to do from here using only the axioms ?!
The only thing you need to prove is that $$(-1)*(-1)=1\tag{1}.$$ If this is done then $$(-a)*(-a)=(-1)*a*(-1)*a=(-1)*(-1)*a*a=1*a*a=a*a.$$
To prove (1), use $$0=(1+(-1))*(-1)=1*(-1)+(-1)*(-1)=-1+(-1)*(-1).$$ So $1=(-1)*(-1)$.