# 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:

 $\displaystyle \left\vert{q_r x - p_r}\right\vert$ $\le$ $\displaystyle \left\vert{b x - a}\right\vert$ 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.

$\blacksquare$