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 $S\mathrm{\ne }\mathrm{\varnothing }$; and $2\mathrm{\notin }S$ since $2>0$ and $2^2=4>2$, so $S\mathrm{\ne }\mathbb{Q}$.

(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 $r={\displaystyle \frac{2q+2}{q+2}}$. Then $r-q={\displaystyle \frac{2-q^2}{q+2}}>0$, so $r>q>0$, while $r^2-2={\displaystyle \frac{2(q^2-2)}{(q+2{)}^2}}<0$, 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 $S\mathrm{\ne }{p}^{\ast }$ for every $p\in \mathbb{Q}$: were $S=p^{\ast }=\{r:r<p\}$, then $p\mathrm{\notin }S$ 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 $s:={\displaystyle \frac{2p+2}{p+2}}$ satisfies $0<s<p$ and $s^2>2$, so $s\mathrm{\notin }S$, 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 $\mathbb{Q}$.