Do the axioms of ordered field imply that $a\cdot 0=0$ and $0<1$?

Question

We introduce some definitions:

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A structure $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.

For all $a,b,c\in A$:

1. $a+b=b+a$.

2. $(a+b)+c=a+(b+c)$.

3. $a+0=a$.

4. There exists $a'\in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.

5. $a<b\implies a+c<b+c$.

6. $a\cdot (b+c)=a\cdot b + a\cdot c$.

7. $a\cdot b = b\cdot a$.

8. $(a\cdot b)\cdot c = a\cdot (b\cdot c)$

9. $a\cdot 1=a$

10. For $a\neq 0$, there exists $a'\in A$ such that $a\cdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.

11. $a<b$ and $0<c$ $\implies a\cdot c < b\cdot c$.

12. $0\neq 1$

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In the familiar structure $\mathfrak{R}=\langle \Bbb R,<,+,\cdot,0,1 \rangle$ where $\Bbb R$ is the set of real numbers, it is trivial that $a\cdot 0=0$ and $0<1$.

I would like to prove that $a\cdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.

Do the axioms of ordered field 1-12 imply that $a\cdot 0=0$ and $0<1$?

Please shed me some light. Thank you for your help!

Answer 1

$a\cdot 0=0$ is a consequence of the basic field axioms; for example $$ a \cdot 0 = a\cdot(0+0) = a\cdot 0 + a\cdot 0 $$ and then cancel one of the $a\cdot 0$s.

For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0\cdot(-1) < (-1)\cdot(-1) = 1$, so $0<1$ after all, a contradiction.

Answer 2

$a \cdot 0=0$ is a consequence of the ring axioms: $a \cdot 0= a \cdot (0+0)= a \cdot 0+ a\cdot 0$.

Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 \cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 \cdot b + (-1)b = 0$, a contradiction.