The rational cuts embed densely in $\mathbb{R}$, preserving sums, products, $0$, $1$ and the order

Order preservation: if $p < q$ then $p^{*} \subsetneq q^{*}$. For $r \in p^{*}$ we have $r < p < q$, so $r \in q^{*}$, giving $p^{*} \subseteq q^{*}$; and $p \in q^{*}$ while $p \notin p^{*}$, so the inclusion is proper.

Order reflection: if $p^{*} \subsetneq q^{*}$ then $p < q$. Pick $r \in q^{*} \setminus p^{*}$; then $r < q$ and $\neg(r < p)$, so $p \le r < q$, whence $p < q$.

Unit identities: $0 \mapsto 0^{*}$ and $1 \mapsto 1^{*}$ hold because $0^{*} = \{\, r < 0 \,\}$ and $1^{*} = \{\, r < 1 \,\}$ are exactly the cuts named by the embedding at $0$ and $1$ and fixed as the additive and multiplicative identities.

Additive identity, inclusion $p^{*} + q^{*} \subseteq (p+q)^{*}$: a typical element is $a + b$ with $a < p$ and $b < q$, and order compatibility of rational addition gives $a + b < p + q$, so $a + b \in (p+q)^{*}$.

Additive identity, inclusion $(p+q)^{*} \subseteq p^{*} + q^{*}$: given $r < p + q$ set $d = (p + q - r)/2 > 0$, $a = p - d$, $b = q - d$; then $a < p$, $b < q$, and $a + b = (p + q) - 2d = r$, so $r = a + b \in p^{*} + q^{*}$.

Nonnegative product, inclusion $p^{*} \cdot q^{*} \subseteq (pq)^{*}$ for $p, q > 0$: an element is either $\le 0$, hence in $(pq)^{*}$ since $pq > 0$, or $ab$ with $0 < a < p$ and $0 < b < q$, and then $ab < pb < pq$, so $ab \in (pq)^{*}$.

Nonnegative product, inclusion $(pq)^{*} \subseteq p^{*} \cdot q^{*}$ for $p, q > 0$: take $r < pq$; if $r \le 0$ it lies in the $\{\, q \le 0 \,\}$ clause, and if $r > 0$ then $r/q < p$, so the strict midpoint $a = (r/q + p)/2$ satisfies $r/q < a < p$, and $b = r/a$ gives $0 < a < p$ and $0 < b < q$ (as $a > r/q > 0$ yields $b = r/a < q$), with $ab = r \in p^{*} \cdot q^{*}$.

Density setup: let $A < B$, i.e. $A \subsetneq B$; choose $x \in B \setminus A$, and since $B$ has no greatest element choose $y \in B$ with $y > x$.

Negation identity $-(p^{*}) = (-p)^{*}$: by the negation definition $-(p^{*}) = \{\, r : \exists\, t > 0,\ -r - t \notin p^{*} \,\} = \{\, r : \exists\, t > 0,\ -r - t \ge p \,\} = \{\, r : \exists\, t > 0,\ r \le -p - t \,\} = \{\, r : r < -p \,\} = (-p)^{*}$, where $-r - t \notin p^{*}$ gives $-r - t \ge p$ by trichotomy and $t = -p - r > 0$ witnesses the last equality.

Additive identity: combining the two inclusions, $(p+q)^{*} = p^{*} + q^{*}$.

Nonnegative multiplicative identity: for $p, q > 0$ the two inclusions give $(pq)^{*} = p^{*} \cdot q^{*}$, while if $p = 0$ or $q = 0$ then $pq = 0$ and the sign rule $0^{*} \cdot B = 0^{*}$ gives $(pq)^{*} = 0^{*} = p^{*} \cdot q^{*}$; hence $(pq)^{*} = p^{*} \cdot q^{*}$ for all $p, q \ge 0$.

Injectivity: if $p^{*} = q^{*}$ then neither $p^{*} \subsetneq q^{*}$ nor $q^{*} \subsetneq p^{*}$, so by reflection $\neg(p < q)$ and $\neg(q < p)$; trichotomy forces $p = q$.

Combining preservation and reflection, $p < q \iff p^{*} \subsetneq q^{*}$, that is $p < q \iff p^{*} < q^{*}$: the embedding preserves and reflects order.

$A < y^{*}$: for $a \in A$, separation gives $a < x$ (as $x \notin A$) and $x < y$, so $a < y$ and $a \in y^{*}$, whence $A \subseteq y^{*}$; and $x \in y^{*}$ (since $x < y$) while $x \notin A$, so the inclusion is proper, $A \subsetneq y^{*}$.

$y^{*} < B$: for $r \in y^{*}$, $r < y$ and $y \in B$, so downward closure gives $r \in B$, whence $y^{*} \subseteq B$; and $y \in B$ while $y \notin y^{*}$, so $y^{*} \subsetneq B$.

Absolute value identity $|p^{*}| = |p|^{*}$: since $0^{*} \subseteq p^{*} \iff$ every $r < 0$ satisfies $r < p \iff p \ge 0$, we have $p \ge 0 \iff p^{*} \ge 0^{*}$; if $p \ge 0$ then $|p^{*}| = p^{*} = |p|^{*}$, while if $p < 0$ then $p^{*} < 0^{*}$, so $|p^{*}| = -(p^{*}) = (-p)^{*} = |p|^{*}$ using $-(p^{*}) = (-p)^{*}$ and $|p| = -p$.

Multiplicative identity for all signs: the sign rules give $p^{*} \cdot q^{*} = \pm(|p^{*}| \cdot |q^{*}|)$, and $|p^{*}| \cdot |q^{*}| = |p|^{*} \cdot |q|^{*} = (|p|\,|q|)^{*}$ by the absolute-value identity and the nonnegative case; when $p, q$ share a sign $pq \ge 0$ and $|p|\,|q| = pq$, so $p^{*} \cdot q^{*} = (pq)^{*}$, and when they have opposite signs $pq < 0$, $|p|\,|q| = -pq$, and $-\bigl((-pq)^{*}\bigr) = (pq)^{*}$ by the negation identity, so again $p^{*} \cdot q^{*} = (pq)^{*}$ (the $p = 0$ or $q = 0$ case being step 2.2); hence $(pq)^{*} = p^{*} \cdot q^{*}$ for all $p, q$.

Taking $q = y$ yields $A < q^{*} < B$, so the image is dense; with injectivity, order preservation/reflection, and the ring identities $(p+q)^{*} = p^{*} + q^{*}$, $(pq)^{*} = p^{*} \cdot q^{*}$, $0 \mapsto 0^{*}$, $1 \mapsto 1^{*}$, the map $q \mapsto q^{*}$ is a dense, order-preserving ring embedding of $\mathbb{Q}$ into $\mathbb{R}$. Closure of the image under reciprocals, which a subfield would also require, is not established here.