Question Number 3768 by Filup last updated on 19/Dec/15
$$\mathrm{let}\:\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{be}\:\mathrm{irrational}. \\ $$$${That}\:{is}: \\ $$$${a},\:{b},\:{c}\in\mathbb{R}/\mathbb{Q} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{for}: \\ $$$${c}=\frac{{a}}{{b}}? \\ $$
Commented by 123456 last updated on 20/Dec/15
$${a}=\sqrt{\mathrm{6}},{b}=\sqrt{\mathrm{3}},{c}=\sqrt{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{\sqrt{\mathrm{6}}}{\sqrt{\mathrm{3}}}=\sqrt{\mathrm{2}}={c} \\ $$