Prove that in an ordered field, $0\cdot x=0$ for every $x$ 
Prove that in an ordered field $0\cdot x=0$ for every $x$.

I'm not sure how to go about proving something is true in an ordered field. What are the steps involved? 
 A: Everyone talking about ordered fields being unnecessary and no one actually helping. Here you go:
Since $0$ is the additive identity, we have $0 = 0 + 0$. So for $x \in F$, we have 
$0\cdot x = (0+0)\cdot x$.
By distributivity, 
$(0+0)\cdot x = 0\cdot x + 0\cdot x$.
Whatever $0\cdot x$ is, it's an element of the field (i.e. ring), so it has an additive inverse $-(0\cdot x)$. Add it to both sides  to obtain
$ 0\cdot x + 0\cdot x + (-(0\cdot x)) = 0\cdot x + (-(0\cdot x))$
$0\cdot x = 0$.
As others have mentioned, this is perfectly true in a regular old ring. You can also prove that $x\cdot 0 = 0$ separately.
A: As zeno's answer shows, the order axioms aren't necessary.  But here's a proof that uses them:

If $0\lt0x$, then
$$0x=0+0x\lt0x+0x=(0+0)x=0x$$
which is a contradiction.  Likewise if $0\gt0x$.  
Therefore $0x=0$.

What's used about ordered fields is:


*

*If $a\lt b$ then $a+c\lt b+c$ for any $c$; and

*For any $a$ and $b$, exactly one of the relations $a\lt b$, $a\gt b$ or $a=b$ holds.


The only other field axioms being used are $0+a=a$ for any $a$ and $ac+bc=(a+b)c$ for any $a,b,c$.
