Exercise 7.6.5* ($\left | \xi - \frac{h}{k} \right| < \frac{1}{2k^2}$)

Proof. (Since the statement doesn’t suppose that $\xi$ is irrational, we treat the general case for any real number $\xi$, irrational or not.)

Suppose for the sake of contradiction that $\left|\xi -\frac{h}{k}\right|<\frac{1}{2k^2}$ is false if we replace $h∕k$ by both $h_n∕k_n$ and $h_{n+1}∕k_{n+1}$ for some $n\ge 0$, so that

By Theorem 7.6, we know that $\xi$ is between $h_n∕k_n$ and $h_{n+1}∕k_{n+1}$, thus

Then (1) and (2) give, using Theorem 7.5,

which is equivalent to

This implies $k_n=k_{n+1}$. Since $k_{n+1}=a_{n+1}{k}_n+k_{n-1}\ge k_n+k_{n-1}$, this is false, except if $k_{n-1}=0$, that is $n=0$, and also $a_{n+1}=a_1=1$.

In this special case ( $n=0,a_1=1$), we obtain $k_1=k_0=1$, $h_0=a_0,h_1=a_0+1$, so $h_0∕k_0=a_0$ and $h_1∕k_1=a_0+1$, where $a_0=\lfloor \xi \rfloor$. Then (1) gives

thus

If $\xi -a_0>1∕2$ or $a_0+1-\xi >1∕2$, then $1>1$, so $\xi -a_0\le 1∕2$ and $a_0+1-\xi \ge \frac{1}{2}$. Hence by (4), $\xi -a_0=1∕2$, so $\xi =⟨a_0,2⟩$, and $h_1∕k_1=\xi$ so $\left|\xi -\frac{h_1}{k_1}\right|=0$ and this contradicts (1).

This contradiction shows that for every $n\ge 0$ such that $h_n∕k_n$ and $h_{n+1}∕k_{n+1}$ are defined,

Of every two consecutive convergents $h_n∕k_n$ to $\xi$ with $n\ge 0$, at least one satisfies

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