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

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

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

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

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

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

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

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

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

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

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

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

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

Combining preservation and reflection, $p<q⟺p^{\ast }⊊q^{\ast }$, that is $p<q⟺p^{\ast }<q^{\ast }$: the embedding preserves and reflects order.

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

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

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

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

Taking $q=y$ yields $A<q^{\ast }<B$, so the image is dense; with injectivity, order preservation/reflection, and the ring identities $(p+q{)}^{\ast }=p^{\ast }+q^{\ast }$, $(pq{)}^{\ast }=p^{\ast }\cdot q^{\ast }$, $0\mapsto 0^{\ast }$, $1\mapsto 1^{\ast }$, the map $q\mapsto q^{\ast }$ 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.