Condition for Rational to be a Convergent

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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:

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

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:

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


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:

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

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.

$\blacksquare$