Are there any "crazy" totally ordered rings without infinities/infinitesimals? I was seeing this post and its nice answer. I asked myself if I could have found a more "crazy" example. I was not very succesful. So I formalized what I was trying to achieve:

Question: Let $(R,+,\cdot,0,1,<)$ be a totally ordered commutative ring without infinities/infinitesimals, i.e.
  $$\forall x\in R:\exists n\in\Bbb N:\underbrace{1+\cdots+1}_{n}>|x|.,
\qquad\forall x\in R-\{0\}:\exists n\in\Bbb N:\underbrace{|x+\cdots+x|}_{n}>1.$$
  Is it true that this already suffices to make $R$ isomorphic to a  subring of $\Bbb R$?

This would show that there are no "crazy" examples. 
My approaches mostly started from $\Bbb R$ by adding new elements, but it seems any such additional element led to infinities/infinitesimals. On the other hand I do not even have an idea on how to show that $R$ might be embedded into $\Bbb R$ in general.

Update: I found the Hahn embedding theorem to be very relevant for my question.
 A: This is indeed true.  Such a totally ordered ring is called Archimedean.  Here's how you embed an Archimedean ordered ring $R$ into $\mathbb{R}$.  For any positive $r\in R$, consider the set $S_r$ of positive rational numbers $\frac{a}{b}$ (with $a,b\in\mathbb{Z}_+$) such that $br\geq a$.  Your second assumption implies $S_r$ is nonempty, and your first assumption implies $S_r$ is bounded above.  Define $f(r)=\sup S_r\in\mathbb{R}$.  Also define $f(0)=0$, and $f(r)=-f(-r)$ if $r<0$.  (Morally, $S_r$ is the set of positive rational numbers less than or equal to $r$, so we are sending $r>0$ to the real number which defines the same cut in the rational numbers as $r$.)
This defines a map $f:R\to\mathbb{R}$, and it is straightforward to check that this map is an order-preserving homomorphism.  It is also injective, since $f(r)$ is by definition always positive if $r>0$ and negative if $r<0$.
(Alternatively, to avoid separate cases based on signs in the definition of $f$, you could define $S_r$ to be the set of $\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and $b>0$ such that $br\geq a$ and define $f(r)=\sup S_r$ for any $r\in R$.  This probably makes it a bit easier to check that $f$ is a homomorphism.  You still have to separate into cases to prove the supremum exists, though: if $r>0$, then $S_r$ is obviously nonempty but you must use the non-existence of infinite elements to prove it is bounded above, while if $r<0$, then $S_r$ is obviously bounded above but you must use the non-existence of infinite elements to prove it is nonempty.  You then have to use the non-existence of infinitesimal elements to prove $f(r)>0$ when $r>0$ and $f(r)<0$ when $r<0$, so that $f$ is injective.)
