Show that $\mathbb{Q} \subset \mathbb{R}$, given the axiomatic definition of $\mathbb{R}$ If instead of constructing $\mathbb{R}$ from $\mathbb{Q}$, we define it axiomatically, as a complete ordered field, how would one show that $\mathbb{Q} \subset \mathbb{R}$?
 A: Well, $1 \in \mathbb{R}$, and it's closed under addition, so $1+1\in \mathbb{R}$. By induction (using the orderness to ensure that we get new elements at each step), $\mathbb{N}\subset \mathbb{R}$. Additive inverses are also present, so $\mathbb{Z}\subset \mathbb{R}$. Now the field of fractions of a ring is the smallest field which contains that ring. And $\mathbb{Q}$ is the field of fractions for $\mathbb{Z}$. And $\mathbb{R}$ is a field containing $\mathbb{Z}$. So $\mathbb{Q}\subset\mathbb{R}$.
A: You can't quite show that. But you can show something close enough.
Because if you define $\Bbb R$ axiomatically, and you claim that $\Bbb R$ is a model of these axioms, then there is no guarantee that the model will coincide with $\Bbb Q$ in any reasonable way.
But you can show that there is an embedding of $\Bbb Q$ into $\Bbb R$ which preserves the ordered field structure (and the image of the embedding is indeed dense in $\Bbb R$).

The point is that $\Bbb Q\subseteq\Bbb R$ is a specific claim about sets, or structures for some language which may or may not be true given the specific sets chosen to interpret these two fields.
But "There is an embedding from $\Bbb Q$, into $\Bbb R$" (however you defined these objects) is a general and abstract statement which is what you really want to prove. What is worth mentioning, as egreg points in the comment, is that such embedding is in fact unique.

To see there is an embedding, proceed by induction using the constants $0,1$ to show that all the natural numbers can be "identified" in $\Bbb R$. Proceed to conclude from the existence of additive inverses that the "identification" extends into $\Bbb Z$. Finally, from multiplicative inverses for non-zero numbers extend this identification to $\Bbb Q$ itself.
We need to show that it preserves addition and multiplication, which can be shown by similar inductions. And the order is preserved by similar arguments of building this from the naturals to the rationals.
A: *

*$1\in\mathbb{R}$

*$n=\underbrace{1+1+...+1}_{\mbox{n - times}}\in\mathbb{R}$

*$-n\in\mathbb{R}$

*$\mathbb{Z}\subset \mathbb{R}$

*Since $\mathbb{R}$ is a field and $\mathbb{Z}\subset \mathbb{R}$ therefore $\frac{p}{q}\in \mathbb{R}$ for all $p\in \mathbb{Z} \wedge q\in \mathbb{Z}\setminus \{0\}$ thus $\mathbb{Q}\subset \mathbb{R}$

