Reduced real rational representations are unique up to scale

For a nonzero quotient, let $d=\gcd (p,q)$ from [F1] and put $p_0=p\mathrm{/}d$, $q_0=q\mathrm{/}d$. Then $Ap+Bq=d$ implies $A{p}_0+B{q}_0=1$, so no nonconstant polynomial divides both $p_0$ and $q_0$. For the zero quotient, $p=0$ and division by the monic $\gcd (0,q)$ gives $(0,\mathrm{lc}(q))$; scaling by $1\mathrm{/}\mathrm{lc}(q)$ gives $(0,1)$. Any reduced pair representing zero has zero numerator and a nonzero constant denominator.

Suppose $p_0\mathrm{/}{q}_0=p_1\mathrm{/}{q}_1$ are nonzero reduced representations. By the stated equality convention, $p_0{q}_1=p_1{q}_0$. The Bézout identity $A{p}_0+B{q}_0=1$ from step 1.1, multiplied by $p_1$, shows $p_0$ divides $p_1$; multiplying the same identity by $q_1$ shows $q_0\mid q_1$. Write $p_1=h{p}_0$ and $q_1=k{q}_0$. The nonzero real leading coefficients of $p_0$ and $q_0$ have nonzero product, so $p_0{q}_0\mathrm{\ne }0$; cross-multiplication then gives $(h-k)p_0{q}_0=0$ and hence $h=k$. If $h$ were nonconstant, it would divide both $p_1$ and $q_1$, contradicting reduction. Thus $h=k$ is a nonzero real constant. The zero case is the reduced pair already identified in step 1.1.

Since $I$ is nonempty, choose $x\in I$. If the common scalar from step 2.1 is $c$, then $q_1(x)=c{q}_0(x)$ and both denominators are positive, so $c>0$. Homogeneity of the coefficient norm and $\mathrm{\parallel }{q}_0{\mathrm{\parallel }}_{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f},2}=\mathrm{\parallel }{q}_1{\mathrm{\parallel }}_{\mathrm{c}\mathrm{o}\mathrm{e}\mathrm{f},2}=1$ give $c=1$, so the numerators agree as well.