Question Number 70885 by naka3546 last updated on 09/Oct/19
$$\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\:=\:{a}\:+\:\frac{\mathrm{1}}{{b}\:+\:\frac{\mathrm{1}}{{c}\:+\:\frac{\mathrm{1}}{{d}\:+\:\ldots}}} \\ $$$${a},\:{b},\:{c},\:{d}\:\:\in\:\mathbb{Z}^{+} \\ $$$${What}'{s}\:\:{b}\:\:? \\ $$
Answered by MJS last updated on 09/Oct/19
$$\mathrm{the}\:\mathrm{continued}\:\mathrm{fraction}\:\mathrm{of}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\mathrm{is}\:\mathrm{non}+\mathrm{periodic} \\ $$$$\Rightarrow\:\mathrm{just}\:\mathrm{calculate}\:\mathrm{it} \\ $$$${a}=\mathrm{1} \\ $$$${b}=\lfloor\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\rfloor=\lfloor\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\rfloor=\mathrm{3} \\ $$