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Let $\alpha$ be a irrational number, then using the continued fraction expansion we can find two sequences $\{p_n\}$ and $\{q_n\}$ with $q_n\rightarrow\infty$ as $n\rightarrow\infty$ such that $$|\alpha-\frac{p_n}{q_n}|<\frac{1}{q_n^2}.$$ My question is that: if now we have two irrational numbers $\alpha$ and $\beta$, can we find sequences $\{a_n\}$, $\{b_n\}$ and $\{q_n\}$ with $q_n\rightarrow\infty$ as $n\rightarrow\infty$ such that $$|\alpha-\frac{a_n}{q_n}|<\frac{1}{q_n^2} \text{ and } |\beta-\frac{b_n}{q_n}|<\frac{1}{q_n^2}?$$ I was trying to get the approximations of $\alpha$ and $\beta$ by rational numbers separately then make the denominators the same but I did not find a way to make it. Maybe this is not true? I am a beginner in the area of diophantine-approximation so I do know if this kind of problems have appeared in somewhere before (I think there should be).

Any idea or reference will be highly appreciated.

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The simultaneous version of Dirichlet's theorem asserts that you can find infinitely many $a_n, b_n$ and $q_n$ such that $$ \left| \alpha - \frac{a_n}{q_n} \right| < \frac{1}{q_n^{3/2}} \; \mbox{ and } \; \left| \beta - \frac{b_n}{q_n} \right| < \frac{1}{q_n^{3/2}}. $$ For almost all pairs of real numbers (including, via a theorem of Schmidt, pairs of independent algebraic numbers), the exponent $3/2$ is best possible.

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  • $\begingroup$ In general, you can simultaneously approximate $n$ irrationals with exponent $(n+1)/n$. $\endgroup$ – user98602 Nov 29 '14 at 5:37

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