Proof that for all Real Numbers $a \in \mathbb R$, $0 \leq a^2$ using certain Axioms

Question

We have the following axioms:

(i) For every $a,b \in \mathbb R$, either $a \leq b$ or $b \leq a$.

(ii) If $a \leq b$ and $b \leq a$, then $a = b$.

(iii) If $a \leq b$ and $b \leq c$, then $a \leq c$.

(iv) If $a \leq b$ and $c \in \mathbb R$, $a + c \leq b + c$

(v)If a $\leq b$ and $0 \leq c$, then $ac \leq bc$.

I want to prove that for every $a \in \mathbb R$, $0 \leq a^2$ using the axioms only. I understand that doing it in two cases will help. But I'm having trouble in the case where I've assumed that $a \leq 0$. I also understand that once I've shown that $0 \leq -a$, then by axiom (v), we have that $a \cdot (-a) \leq 0 \cdot (-a)$, implying that $-a^2 \leq 0$.

My question is, how do you prove that $0 \leq -a$ using only the axioms above, and likewise for proving that $0 \leq a^2$ from $-a^2 \leq 0$?

Thanks in advance.

Answer 1

Let $a\in\mathbb{R}$ with $a\leq 0$. Then $(-a)\in\mathbb{R}$ and axiom (IV) gives us:

$a+(-a)\leq 0+(-a)$ which is (by the field axioms) $0\leq -a$, what you wanted to show.

Answer 2

You use axiom (iv).

With $b=0$ and $c=-a$: $a\le0\implies a+-a\le0+-a\implies 0\le -a$.

Substitute $-a^2$ for $a$ and $0$ for $b$ and $a^2$ for $c$: $-a^2\le0\implies -a^2+a^2\le 0+a^2\implies 0\le a^2$.