# Approximate irrational numbers with the same denominator

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.

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