Incorrect. $\displaystyle\frac{{3}}{{5}}\usepackage{cmbright}\Huge\$ is equivalent to $\displaystyle\frac{{9}}{{15}}\usepackage{cmbright}\Huge\$ , and because $\displaystyle{9}<{10}\usepackage{cmbright}\Huge\$ , $\displaystyle\frac{{9}}{{15}}<\frac{{10}}{{15}}\usepackage{cmbright}\Huge\$ , which means that $\displaystyle\frac{{3}}{{5}}<\frac{{10}}{{15}}\usepackage{cmbright}\Huge\$ .