Condition for Rational to be Convergent

Theorem
Let $x$ be an irrational number.

Let the rational number $\dfrac a b$ satisfy the inequality:
 * $\left\vert{x - \dfrac a b}\right\vert < \dfrac 1 {2 b^2}$

Then $\dfrac a b$ is a convergent of $x$.

Proof
Suppose to the contrary, that $\left\vert{x - \dfrac a b}\right\vert < \dfrac 1 {2 b^2}$ but that $\dfrac a b$ is not one of the convergents $\dfrac {p_n} {q_n}$ of $x$.

Let $r$ be the unique integer for which $q_r \le b \le q_{r+1}$.

Then:

Therefore:
 * $\displaystyle q_r \left\vert{x - \frac {p_r} {q_r}}\right\vert < \frac 1 {2b}$

and so:
 * $\displaystyle \left\vert{x - \frac {p_r} {q_r}}\right\vert < \frac 1 {2 q_r b}$

Hence:

Now note that $q_r a - p_r b$ is a integer, and also non-zero otherwise $\dfrac a b = \dfrac {p_r} {q_r}$ and we supposed (at the top of this proof) that it's not.

But we have:

So, combining results $(1)$ and $(2)$, we get:
 * $\displaystyle \frac 1 {q_r b} < \frac 1 {2 q_r b} + \frac 1 {2b^2}$

This simplifies to $q_r > b$, which contradicts our initial assumptions.