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

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

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

$$\frac{4}{11}<\frac{x}{y}<\frac{3}{8}$$ $$\frac{8x}{3}<y<\frac{11x}{4}$$ $$\lfloor \frac{8x}{3}\rfloor +1⩽y⩽\lceil \frac{11x}{4}\rceil -1$$ $$\lfloor \frac{8x}{3}\rfloor +1⩽\lceil \frac{11x}{4}\rceil -1$$ $$\Rightarrow x_{min}=7$$ $$\Rightarrow y_{min}=19$$ $${(x+y)}_{min}=7+19=26$$ $$$$ $$thenextonesare:$$ $$\frac{7}{19},\frac{10}{27},\frac{11}{30},\frac{13}{35},\frac{16}{43},...$$

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