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 $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ with $b', d' > 0$, i.e. $ab' = a'b$ and $cd' = c'd$; 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 $b'd' > 0$: $(ad)(b'd') \le (cb)(b'd')$.

Rearranging both sides with $ab' = a'b$ and $cd' = c'd$: $(ad)(b'd') = (ab')(dd') = (a'b)(dd') = (a'd')(bd)$ and $(cb)(b'd') = (cd')(bb') = (c'd)(bb') = (c'b')(bd)$.

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: $(a'd')(bd) \le (c'b')(bd)$ with $bd > 0$, so $a'd' \le c'b'$ 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.