Each rational cut $q^{*}$ is a Dedekind cut

(C1) $q^{*}$ is nonempty and proper: $q - 1 < q$ gives $q - 1 \in q^{*}$, while $q \not< q$ gives $q \notin q^{*}$, so $q^{*} \ne \emptyset$ and $q^{*} \ne \mathbb{Q}$.

(C2) $q^{*}$ is downward closed: if $p \in q^{*}$, so $p < q$, and $r < p$, then $r < q$ by transitivity, hence $r \in q^{*}$.

(C3) $q^{*}$ has no greatest element: if $p \in q^{*}$ then $p < q$, so the midpoint $m = \tfrac{p+q}{2}$ satisfies $p < m < q$, giving $m \in q^{*}$ with $m > p$.

Satisfying (C1), (C2), (C3), $q^{*}$ is a Dedekind cut; applied at $q = 0$ and $q = 1$ this shows $0^{*}$ and $1^{*}$ are Dedekind cuts and hence elements of $\mathbb{R}$.