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Question Number 75033 by naka3546 last updated on 06/Dec/19

(4/(11)) < (x/y) < (3/8)  x, y  ∈  Z^+   min {x+y}  =  ?

$$\frac{\mathrm{4}}{\mathrm{11}}\:<\:\frac{{x}}{{y}}\:<\:\frac{\mathrm{3}}{\mathrm{8}} \\ $$ $${x},\:{y}\:\:\in\:\:\mathbb{Z}^{+} \\ $$ $${min}\:\left\{{x}+{y}\right\}\:\:=\:\:? \\ $$

Answered by mr W last updated on 06/Dec/19

(4/(11))<(x/y)<(3/8)  ((8x)/3)<y<((11x)/4)  ⌊((8x)/3)⌋+1≤y≤⌈((11x)/4)⌉−1  ⌊((8x)/3)⌋+1≤⌈((11x)/4)⌉−1  ⇒x_(min) =7  ⇒y_(min) =19  (x+y)_(min) =7+19=26    the next ones are:  (7/(19)), ((10)/(27)), ((11)/(30)), ((13)/(35)), ((16)/(43)), ...

$$\frac{\mathrm{4}}{\mathrm{11}}<\frac{{x}}{{y}}<\frac{\mathrm{3}}{\mathrm{8}} \\ $$ $$\frac{\mathrm{8}{x}}{\mathrm{3}}<{y}<\frac{\mathrm{11}{x}}{\mathrm{4}} \\ $$ $$\lfloor\frac{\mathrm{8}{x}}{\mathrm{3}}\rfloor+\mathrm{1}\leqslant{y}\leqslant\lceil\frac{\mathrm{11}{x}}{\mathrm{4}}\rceil−\mathrm{1} \\ $$ $$\lfloor\frac{\mathrm{8}{x}}{\mathrm{3}}\rfloor+\mathrm{1}\leqslant\lceil\frac{\mathrm{11}{x}}{\mathrm{4}}\rceil−\mathrm{1} \\ $$ $$\Rightarrow{x}_{{min}} =\mathrm{7} \\ $$ $$\Rightarrow{y}_{{min}} =\mathrm{19} \\ $$ $$\left({x}+{y}\right)_{{min}} =\mathrm{7}+\mathrm{19}=\mathrm{26} \\ $$ $$ \\ $$ $${the}\:{next}\:{ones}\:{are}: \\ $$ $$\frac{\mathrm{7}}{\mathrm{19}},\:\frac{\mathrm{10}}{\mathrm{27}},\:\frac{\mathrm{11}}{\mathrm{30}},\:\frac{\mathrm{13}}{\mathrm{35}},\:\frac{\mathrm{16}}{\mathrm{43}},\:... \\ $$

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