Condition for Rational to be a Convergent

Theorem
Let $x$ be an irrational number.

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

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

Proof
$\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, 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:
 * $q_r \size {x - \dfrac {p_r} {q_r} }< \dfrac 1 {2 b}$

and so:
 * $\size {x - \dfrac {p_r} {q_r} } < \dfrac 1 {2 q_r b}$

Hence:

Now note that $q_r a - p_r b$ is a integer.

However, $\dfrac a b$ is not one of the convergents $\dfrac {p_n} {q_n}$ of $x$.

Thus $\dfrac a b \ne \dfrac {p_r} {q_r}$.

But we have:

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

This simplifies to $q_r > b$.

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

From this contradiction it follows that $\dfrac a b$ is one of the convergents of $x$