Prove $\mathbb{Q}$ is dense without reference to $\mathbb{R}$

Question

If $x, y \in \mathbb{R}$ and $x < y$, then there exists a $\frac p q \in \mathbb{Q}$ such that $x < \frac p q < y$. To prove this, it's sufficient to find integers $q \neq 0$ and $p$ such that $qx < p < qy$, and this is true by Archimedean property: there exists $q \in \mathbb{N}$ such that $q(y - x) > 1$, i.e. $qy > qx + 1$.

Since $\mathbb{Q}$ is a subfield of $\mathbb{R}$, the same result holds if $x, y \in \mathbb{Q}$. But how can we prove it without constructing $\mathbb{R}$?

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The motivation: if we have this, then it's easy to show that $\mathbf{0} = \{y \in \mathbb{Q} : y < 0 \}$ is a Dedekind cut; in particular, it's easy to show that $\mathbf{0}$ contains no maximum. Then we can let $\mathbf{0}$ be the identity for the set of Dedekind cuts under (closed) set addition, as discussed in Dedekind Cuts - Additive and Multiplicative Identities.

Answer 1

If $x, y \in \mathbb{Q}$, and $x < y$, then $\frac{x+y}{2} \in \mathbb{Q}$ and $x < \frac{x+y}{2} < y$.

Answer 2

Any topological space is dense in itself, so this is trivially true. I guess you meant instead to ask about the following:

If $x, y \in \Bbb{Q}$ with $x < y$, then there is $r\in\Bbb{Q}$ with $x <r < y$ and yes, this is also trivial. Simply take $$r:= \frac{x+y}{2}\in \Bbb{Q}$$