The cut $S = \{q : q<0 \text{ or } q^2<2\}$ is an irrational real number

(C1) $0 \in S$ since $0^2 = 0 < 2$, so $S \ne \emptyset$; and $2 \notin S$ since $2 > 0$ and $2^2 = 4 > 2$, so $S \ne \mathbb{Q}$.

(C2) Let $p \in S$ and $q < p$. If $q < 0$ then $q \in S$ by definition. Otherwise $0 \le q < p$, so $p > 0$; then $p \in S$ forces $p^2 < 2$, and $0 \le q < p$ gives $q^2 < p^2 < 2$, whence $q \in S$.

(C3, case $q \le 0$) Given $q \in S$ with $q \le 0$, take $r = 1$: then $1 \in S$ (as $1^2 = 1 < 2$) and $r = 1 > 0 \ge q$, so $r > q$.

(C3, case $q > 0$) Given $q \in S$ with $q > 0$, we have $q^2 < 2$; set $r = \dfrac{2q+2}{q+2}$. Then $r - q = \dfrac{2 - q^2}{q+2} > 0$, so $r > q > 0$, while $r^2 - 2 = \dfrac{2\,(q^2 - 2)}{(q+2)^2} < 0$, so $r^2 < 2$; hence $r \in S$ with $r > q$.

(C3) Combining the two cases, every $q \in S$ admits $r \in S$ with $r > q$: $S$ has no greatest element.

By steps 1.1, 1.2 and 2.1, $S$ satisfies (C1)–(C3); it is a Dedekind cut (Dedekind cut), i.e. a real number (The real numbers $\mathbb{R}$ as Dedekind cuts).

Finally $S \ne p^\ast$ for every $p \in \mathbb{Q}$: were $S = p^\ast = \{ r : r < p \}$, then $p \notin S$ would give $p > 0$ and $p^2 \ge 2$, while $p^2 = 2$ is impossible by [L2], so $p^2 > 2$; as $p > 0$ we have $p + 2 > 0$, and then $s := \dfrac{2p+2}{p+2}$ satisfies $0 < s < p$ and $s^2 > 2$, so $s \notin S$, yet $s < p$ puts $s \in p^\ast = S$, a contradiction. Thus $S$ is a cut represented by no rational: it is the cut that names $\sqrt{2}$, the real number absent from $\mathbb{Q}$.