Exercise 7.5.1 (The first assertion in Theorem 7.13 holds in case $n = 0$ if $k_1>1$)

Proof. Assume that $k_1>1$. By (7.6) $k_1=a_1$, thus $a_1>1$.

We prove that the first assertion of Theorem 7.13 holds in case $n=0$. We must show that for all fractions $a∕b$ with $b>0$,

Suppose at the contrary that there exists some fraction $a∕b$ with $0<b\le 1$ such that $\left|\xi -\frac{a}{b}\right|<\left|\xi -\frac{h_0}{k_0}\right|$. Then $b=1$, and $h_0∕k_0=a_0$ by (7.6), so

Since $a_0=\lfloor \xi \rfloor$ by (7.7), $0\le \xi -a_0<1$, thus $\left|\xi -a\right|<1$ by (2). Hence

so $a_0-1<a<a_0+2$. Since $a$ and $a_0$ are integers, we obtain $a=a_0$ or $a=a_0+1$. But (2) shows that $a\ne a_0$, so

Then $\xi -a=\xi -a_0-1<0$, so (2) becomes $a-\xi <\xi -a_0$, that is $a_0+1-\xi <\xi -a_0$, therefore

Then $1∕2<\xi -a_0<1$, so $1<\frac{1}{\xi -a_0}<2$. Using (7.7), we obtain

Since $k_1=a_1>1$ by hypothesis, this is a contradiction, so (1) is true.

In conclusion, if $k_1>1$ and $a∕b$ is a rational number with positive denominator such that $|\xi -a∕b|<|\xi -h_0∕k_0|$, then $b>k_0=1$. □