Exercise 7.5.2 (Theorem 7.13 becomes false if $bk_n$ is replaced by $b \geq k_{n+1}$)

Question

Prove that the first assertion of Theorem 7.13 becomes false if $b>k_n$ is replaced by $b \geq k_{n+1}$.

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Hint. Use $\xi = \pi^{-1}$ and $n = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Put 
$$\xi = \pi^{-1} = \langle 0, 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, \ldots \rangle.$$
The first convergents are
$$r_0 = 0, r_1 = 1/3, r_2 = 7/22, r_3 = 106/33, r_4 = 113/355.$$
Then $r_1 = h_1/k_1 = 1/3$, $k_2 = 22$.
Take $a = 4,\ b = 13$. Then
$$\left | \frac{1}{\pi} - \frac{4}{13} \right | < 0.0107 < 0.0150 < \left | \frac{1}{\pi} - \frac{1}{3} \right | ,$$
and $b \leq 22 = k_2$.

(We can take also $a/b = 5/16$, or $a/b = 6/19$.)

This shows that there exists a rational number $a/b$ with positive denominator such that $|\xi - a/b| < |\xi - h_1/k_1|$, and $b < k_2$. So the first assertion of Theorem 7.13 becomes false if $b>k_n$ is replaced by $b \geq k_{n+1}$.
\end{proof}

Exercise 7.5.2 (Theorem 7.13 becomes false if $b>k_n$ is replaced by $b \geq k_{n+1}$)

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Exercise 7.5.2 (Theorem 7.13 becomes false if $b>k_n$ is replaced by $b \geq k_{n+1}$)

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