Question Number 12405 by Ms.Ramanujan last updated on 21/Apr/17
$${how}\:{to}\:{prove}\:{that}\:{there}\:{exist}\:{infinitely}\:{many}\:{rationals}\:{between}\:{any}\:{two}\:{irrationals}? \\ $$$$ \\ $$
Commented by FilupS last updated on 21/Apr/17
$${a}<{x}<{b} \\ $$$$\: \\ $$$$\mathrm{let}\:{x}=\:\frac{{b}}{{na}}\:\:\:\:{a},{b},{n}\in\mathbb{Z} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{an}\:\mathrm{infinte}\:\mathrm{number}\:\mathrm{of}\:\mathrm{rationals} \\ $$$$\mathrm{between}\:{a}\:{and}\:{b} \\ $$$$\: \\ $$$$\mathrm{let}\:\frac{{b}}{{a}}\in\mathbb{R}\backslash\mathbb{Q},\:{n}\in\mathbb{R} \\ $$$${x}=\frac{{b}}{{na}}\in\mathbb{Q}\:\:\:\:\:\mathrm{e}.\mathrm{g}.\:\:{x}=\frac{\mathrm{1}}{{k}\left(\frac{\mathrm{1}}{\pi}\right)×\pi}\:\:\:\:\:\:\:{k}\in\mathbb{Q} \\ $$$$\:\:\:\:\:\:\mathrm{or}\:\:{x}=\frac{\pi+\mathrm{1}}{{n}\pi}\:\Rightarrow\:{n}={k}\left(\mathrm{1}+\frac{\mathrm{1}}{\pi}\right) \\ $$$$\: \\ $$$$\therefore\forall{a}\forall{b}\in\left\{\mathbb{R}\backslash\mathbb{Q}\right\}\exists\left\{{x}\right\}\in\left({a},{b}\right):{x}\in\mathbb{Q}\wedge\mid\left\{{x}\right\}\mid=\infty \\ $$