The rationals form a totally ordered field

Order-scaling in $\mathbb{Z}$: for $p>0$, if $u<v$ then $0<(v-u)p$, so $up<vp$; conversely if $up\le vp$ and $v<u$ then $vp<up$, impossible; hence $u\le v$ iff $up\le vp$.

Suppose $(a,b)\sim (a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ and $(c,d)\sim (c^{\mathrm{\prime }},d^{\mathrm{\prime }})$ with $b^{\mathrm{\prime }},d^{\mathrm{\prime }}>0$, i.e. $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$ and $c{d}^{\mathrm{\prime }}=c^{\mathrm{\prime }}d$; suppose also $ad\le cb$.

Totality: $ad\le cb$ or $cb\le ad$ in $\mathbb{Z}$, so $x\le y$ or $y\le x$.

Antisymmetry: $ad\le cb$ and $cb\le ad$ give $ad=cb$, i.e. $x=y$ as classes.

Positive products: if $0<x$ and $0<y$ then $0<a$ and $0<c$, so $xy=[(ac,bd)]$ has $ac>0$ and $bd>0$, hence $0<xy$.

For transitivity, let $z=[(e,f)]$ with $f>0$ and suppose additionally $y\le z$, i.e. $cf\le ed$.

Scaling the hypothesis $ad\le cb$ by $b^{\mathrm{\prime }}{d}^{\mathrm{\prime }}>0$: $(ad)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})\le (cb)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})$.

Rearranging both sides with $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$ and $c{d}^{\mathrm{\prime }}=c^{\mathrm{\prime }}d$: $(ad)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})=(a{b}^{\mathrm{\prime }})(d{d}^{\mathrm{\prime }})=(a^{\mathrm{\prime }}b)(d{d}^{\mathrm{\prime }})=(a^{\mathrm{\prime }}{d}^{\mathrm{\prime }})(bd)$ and $(cb)(b^{\mathrm{\prime }}{d}^{\mathrm{\prime }})=(c{d}^{\mathrm{\prime }})(b{b}^{\mathrm{\prime }})=(c^{\mathrm{\prime }}d)(b{b}^{\mathrm{\prime }})=(c^{\mathrm{\prime }}{b}^{\mathrm{\prime }})(bd)$.

Transitivity: from $ad\le cb$ and $cf\le ed$, scaling by $f>0$ and $b>0$ gives $(af)d=(ad)f\le (cb)f=(cf)b\le (ed)b=(eb)d$; cancelling $d>0$ via order-scaling, $af\le eb$, i.e. $x\le z$.

Compatibility with addition: $x+z\le y+z$ reads $(af+eb)(df)\le (cf+ed)(bf)$, which expands to $(ad)f^2+(eb)(df)\le (cb)f^2+(ed)(bf)$; the second terms are equal, so this is $(ad)f^2\le (cb)f^2$, equivalent by order-scaling with $f^2>0$ to $ad\le cb$, i.e. $x\le y$.

Combining: $(a^{\mathrm{\prime }}{d}^{\mathrm{\prime }})(bd)\le (c^{\mathrm{\prime }}{b}^{\mathrm{\prime }})(bd)$ with $bd>0$, so $a^{\mathrm{\prime }}{d}^{\mathrm{\prime }}\le c^{\mathrm{\prime }}{b}^{\mathrm{\prime }}$ by order-scaling: the order is independent of representatives.

The order is well defined, total, compatible with addition, and positives are closed under multiplication: $\mathbb{Q}$ is a totally ordered field.