Prove using the axioms that the square of any number is nonnegative How do you prove $\forall x\in \Bbb{R}, x^2 \ge 0$ using the axioms?



My lecturer hinted you should split the cases up into $x=0$ and $x \ne 0$.
The $x=0$ case seems pretty obvious: $x^2 =x \cdot x = 0 \cdot 0 = 0$ (however I can't think of any axioms to justify this??)
I don't know how to approach the $x \ne 0$ case.

According to trichotomy $O_1$, you have either $x >0$ or $x<0$.
If $x >0$, then according to $O_4$ you get $x^2>0$ as desired.
And if $x <0$ then using $O_3$ you get $x + (-x) < -x$, i.e. $0 < -x$.
Using what is above $ 0 < (-x) \cdot (-x)$.
With $F_9$
$$0 = (-x)0 = (-x)(-x + x)= (-x)(-x) + (-x)x = (-x)(-x) + (-1)x^2 = (-x)(-x) + -x^2$$
Adding $x^2$ on both sides of the equality your get $x^2 = (-x)(-x)$ by use of $F_4$.
Plugging that in the inequality $ 0 < (-x) \cdot (-x)$ you conclude again that $0 < x^2$.
