Exercise 7.5.4 ( Every ``good approximation'' to $\xi$ is a convergent.)

Proof. Here $\xi$ is an irrational number.

Let $a∕b$ be a “good approximation” to $\xi$. By definition $b>0$ and

so

We show first that if $y\ne b$, then the inequality in (1) is strict.

Assume, for the sake of contradiction, that $0<y<b$ and $|\mathit{\xi b}-a|\le |\mathit{\xi y}-x|$. Then $\mathit{\xi b}-a=𝜀(\mathit{\xi y}-x)$, where $𝜀=\pm 1$.

• If $𝜀=-1$, then $\mathit{\xi b}-a=-\mathit{\xi y}+x$, thus $\xi (b+y)=a+x$, where $b+y>0$, hence $\xi =\frac{a+x}{b+y}\in \mathbb{Q}$. This contradicts our hypothesis.
• If $𝜀=+1$, then $\mathit{\xi b}-a=\mathit{\xi y}-x$, thus $\xi (b+y)=a+c$, thus $\xi (y-b)=x-a$, where $y<b$, hence $\xi =\frac{x-a}{y-b}$, which is also a contradiction.

This shows that

Since the sequence ${(k_n)}_{n\in \mathbb{N}}$ is strictly increasing and $k_0=1,b>0$, there exists some integer $n\in \mathbb{N}$ such that $k_n\le b<k_{n+1}$. We want to show that $b=k_n$ and $a=h_n$.

If $b\ne k_n$, then $0\le k_n<b$ and by (2),

Then the second part of Theorem 7.13 shows that $b\ge k_{n+1}$, and this contradicts the definition of $n$. Therefore $b=k_n$.

By (1), $|\mathit{\xi b}-a|\le |\xi k_n-h_n|$. If $|\mathit{\xi b}-a|<|\xi k_n-h_n|$, then by the second part of Theorem 7.13 $b\ge k_{n+1}$, which is false, so $|\mathit{\xi b}-a|=|\xi k_n-h_n|$, where $b=k_n$, so

Therefore $\xi k_n-a=𝜀(\xi k_n-h_n)$, where $𝜀=\pm 1$.

If $𝜀=-1$, then $2\xi k_n=a+h_n$, where $k_n>0$, thus $\xi =(a+h_n)∕(2k_n)\in \mathbb{Q}$. This is a contradiction. Therefore $𝜀=1$, so $\xi k_n-a=\xi k_n-h_n$ and $a=h_n$.

This shows that $a∕b=h_n∕k_n$ is a convergent.

Every “good approximation” to $\xi \in ℝ\setminus \mathbb{Q}$ is a convergent. □