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

Question

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?

That the field is ordered is irrelevant. Even that it's a field isn't important, that holds in every ring. — Daniel Fischer, Sep 06 '16 at 19:24
One of the answers to "Fields - Proof that every multiple of zero equals zero" contains a proof that works just fine. — , Sep 06 '16 at 19:55
Please don't use $$ for multiplication in mathematics. It usually signifies convolution*, which is another beast entirely. (I fixed it for you.) — Harald Hanche-Olsen, Sep 06 '16 at 20:00

zeno · Answer 1 · 2016-09-06T20:09:04.180

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.

Barry Cipra · Answer 2 · 2016-09-06T20:44:15.590

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 $a,b,c$.

2 Answers2