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Let $a$ be a positive irrational number. Let $p_k/q_k, p_{k+1}/q_{k+1}$ be two consecutive convergents of its simple continued fraction, where $k\ge 1$.

Is it possible that both $$|a-(p_k/q_k)|<1/(2q_k^2)$$ and $$|a-(p_{k+1}/q_{k+1})|<1/(2q_{k+1}^2)$$ are true?

I can only prove that at least one of these inequalities is true.

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Do you have $\lfloor a \rfloor = p_0/q_0$? –  Aryabhata Mar 9 '11 at 15:27
    
Yes. That is right. –  TCL Mar 9 '11 at 16:32

2 Answers 2

At least it can happen that not both are true. Example: 333/106 is a convergent to $\pi$, but $(\pi-(333/106)) \cdot 2 \cdot 106^2 \approx 1.87 > 1$.

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+1: Verified with a calculator :-) –  Aryabhata Mar 9 '11 at 15:33
    
But it can also happen for some $a$ and $k$ that both are true. N[Map[2 (# - Pi) Denominator[#]^2 &, Convergents[Pi,30]],10] in Mathematica gives a list of numbers which vary somewhat randomly between being $<1$ or $>1$ in absolute value. –  Hans Lundmark Mar 9 '11 at 15:36
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...and there are places where several consecutive values are $<1$ (but no two consecutive values are $>1$, which agrees with what you claim to have proved). –  Hans Lundmark Mar 9 '11 at 15:40
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Of any two consecutive, at least one of them satisfies the condition in the question. That is true. It is an exercise in the Number Theory book by Niven. –  Aryabhata Mar 9 '11 at 15:50

It is known that $$|a-(p_k/q_k)|\le1/(q_kq_{k+1})\le1/(a_{k+1}q_k^2)$$ (see, e.g., Hardy and Wright) so if all the partial quotients exceed 2 then all the convergents satisfy your inequalitites.

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