# How to prove that $\frac 1{q_{n+2}}\leq \left|p_{n}-bq_n\right|$?

We know that \begin{equation*} a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cfrac{1}{\ddots+\cfrac{1}{a_n}}}}}=[a_0,a_1, \cdots, a_n] \end{equation*}

If $\frac{p_n}{q_n}=[a_0,a_1, \cdots, a_n]$.

To prove that $$\frac 1{q_{n+2}}\leq \left|p_{n}-bq_n\right|\leq \frac 1{q_{n+1}}$$

Now I know that The consecutive convergents are alternately greater than and less than the value of $b$. Then we have that

$${1\over q_nq_{n+1}}=\left|{p_{n+1}q_n-p_nq_{n+1}\over q_nq_{n+1}}\right|=\left|{p_{n+1}\over q_{n+1}}-{p_n\over q_n}\right|=\left|{p_{n+1}\over q_{n+1}}-b\right|+\left|{p_{n}\over q_{n}}-b\right|$$

The last equality follows because the one is bigger and the other is smaller than $b$; this is where it is essential that they are consecutive convergents.

This implies $$\left|p_{n}-bq_n\right|\leq \frac 1{q_{n+1}}$$

How to prove that $$\frac 1{q_{n+2}}\leq \left|p_{n}-bq_n\right|$$?

$$\left|{p_{n}\over q_{n}}-b\right|=\left|{p_{n+1}\over q_{n+1}}-{p_n\over q_n}\right|-\left|{p_{n+1}\over q_{n+1}}-b\right|\Rightarrow \left|{p_{n}\over q_{n}}-b\right|>{1\over q_nq_{n+1}}-{1\over q_{n+1}q_{n+2}}={a_{n+2}\over q_nq_{n+2}}>{1\over q_nq_{n+2}}$$