Condition for Rational to be Convergent

Theorem
Let $$x$$ be an irrational number.

Let the rational number $$\frac a b$$ satisfy the inequality:
 * $$\left|{x = \frac a b}\right| < \frac 1 {2 b^2}$$.

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

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

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

Then:

$$ $$ $$ $$

Therefore $$q_r \left|{x - \frac {p_r} {q_r}}\right| < \frac 1 {2b}$$, and so $$\left|{x - \frac {p_r} {q_r}}\right| < \frac 1 {2 q_r b}$$.

$$ $$

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

But we have:

$$ $$

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

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