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Differentiating $R=a/q+bq$ gives $R^{\prime }=-a/q^2+b$, so for positive stratification its minimum has $q_m^2=a/b$. At that pointThus the two sinks are equal. The buoyancy fraction of total turbulent sinks is always $B/(B+ϵ)=1/2$ there. But the Flux Richardson number defined in part (i) is insteadIt equals $1/2$ only when the minimum is also an equilibrium, $R_{\min }=1$, namely at the critical gradient. This is the qualification needed for the printed assertion; an arbitrary minimum on a curve plotted away from equilibrium need not have ${\mathrm{Ri}}_f=1/2$.

For example choose closure parameters $C_u=C_{\rho }=1/4$, units with $L_u=L_{\rho }=S=1$, and $N^2=1/16$. Then $a=1/64$, $b=4$, $q_m=1/16$ and $R_{\min }=1/2$. The actual flux Richardson number at this minimum is $1/4$, while the buoyancy fraction of the total sinks is $1/2$. This is a concrete counterexample to the unqualified wording. With zero mass density gradient there is no positive interior minimum at all, and ${\mathrm{Ri}}_f=0$.