Condition for Rational to be a 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{q_r x - p_r}\right\vert\)	\(\le\)	\(\displaystyle \left\vert{b x - a}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Convergents are Best Approximations
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b \left\vert{x - \frac a b}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle b \times \frac 1 {2 b^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {2b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac a b - \frac {p_r} {q_r} }\right\vert\)	\(\le\)	\(\displaystyle \left\vert{x - \frac {p_r} {q_r} }\right\vert + \left\vert{x - \frac a b}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Triangle Inequality
\((1):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \frac 1 {2 q_r b} + \frac 1 {2b^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\frac a b - \frac {p_r} {q_r} }\right\vert\)	\(=\)	\(\displaystyle \left\vert{\frac {q_r a - p_r b} {q_r b} }\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\((2):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\ge\)	\(\displaystyle \frac 1 {q_r b}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.

$\blacksquare$