Why is the additive group of an ordered division ring divisible, with a dense underlying set?

Question

I'm reading tame topology and o minimality by Van Den Dries. He defines an ordered ring as follows: a ring with 1 st:

• $0<1$
• $x<y \implies x+z<y+z$ for any $z$
• $x<y$ and $0<z \implies xz<yz$.

He then claims that if the ordered ring is also a division ring, that then the additive group of the ring is divisible, and that the underlying set of the ring is dense ($x<y \implies$ there exist $z$ st $x<z<y$).

Answer 1

Divisible:
Take an $x$ in the field and a natural number $n$. By the first two axioms and induction, $n>0$, so $n\neq0$ and therefore $\frac1n$ exists in our ring. Now consider $\frac1n\cdot x$.

Dense:
Take $x<y$ in the ring. Now consider $z=\frac12(y-x)>0$ (existence and positivity of $z$ must be shown, of course). We have $$ 0<z\implies x<x+z\\ -z<0\implies y-z<y $$ Since $x+z=y-z$, density follows.

Answer 2

Let's say $D$ is an ordered division ring. Notice the axioms imply $D$ has characteristic zero, so $\mathbb{Q}\subseteq D$. To show $(D,+)$ is divisible, we have to show that each $a\in A$ and $n\geq 1$ admits some $x\in D$ so that $nx=a$, but this is easy with $x:=a/n$.