Question Number 181489 by Shrinava last updated on 25/Nov/22 | ||
$$\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{are}\:\mathrm{mutuallly}\:\mathrm{prime}\:\mathrm{numbers} \\ $$ $$\mathrm{If}\:\:\:\frac{\mathrm{3}}{\mathrm{8}}\:<\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}\:<\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $$\mathrm{Find}\:\:\:\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}\right)_{\boldsymbol{\mathrm{min}}} \\ $$ | ||
Answered by mr W last updated on 26/Nov/22 | ||
$$\frac{\mathrm{3}}{\mathrm{8}}<\frac{{a}}{{b}}<\frac{\mathrm{2}}{\mathrm{5}} \\ $$ $$\frac{\mathrm{3}{b}}{\mathrm{8}}<{a}<\frac{\mathrm{2}{b}}{\mathrm{5}} \\ $$ $$\lfloor\frac{\mathrm{3}{b}}{\mathrm{8}}\rfloor+\mathrm{1}\leqslant{a}\leqslant\lceil\frac{\mathrm{2}{b}}{\mathrm{5}}\rceil−\mathrm{1} \\ $$ $$\lfloor\frac{\mathrm{3}{b}}{\mathrm{8}}\rfloor+\mathrm{1}\leqslant\lceil\frac{\mathrm{2}{b}}{\mathrm{5}}\rceil−\mathrm{1} \\ $$ $$\lceil\frac{\mathrm{2}{b}}{\mathrm{5}}\rceil−\lfloor\frac{\mathrm{3}{b}}{\mathrm{8}}\rfloor\geqslant\mathrm{2} \\ $$ $$\Rightarrow\frac{{a}}{{b}}=\frac{\mathrm{5}}{\mathrm{13}},\frac{\mathrm{7}}{\mathrm{18}},\frac{\mathrm{8}}{\mathrm{21}},\frac{\mathrm{9}}{\mathrm{23}},\frac{\mathrm{11}}{\mathrm{28}},\frac{\mathrm{11}}{\mathrm{29}},\frac{\mathrm{12}}{\mathrm{31}},... \\ $$ $$\left({b}−{a}\right)_{{min}} =\mathrm{13}−\mathrm{5}=\mathrm{8} \\ $$ | ||
Commented byShrinava last updated on 28/Nov/22 | ||
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$ | ||