# 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

Aiming for a contradiction, suppose $\size {x - \dfrac a b} < \dfrac 1 {2 b^2}$, 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:

 $\ds \size {q_r x - p_r}$ $\le$ $\ds \size {b x - a}$ Convergents are Best Approximations $\ds$ $=$ $\ds b \size {x - \frac a b}$ $\ds$ $<$ $\ds b \times \frac 1 {2 b^2}$ $\ds$ $=$ $\ds \frac 1 {2 b}$

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:

 $\ds \size {\frac a b - \frac {p_r} {q_r} }$ $\le$ $\ds \size {x - \frac {p_r} {q_r} } + \size {x - \frac a b}$ Triangle Inequality $\text {(1)}: \quad$ $\ds$ $<$ $\ds \frac 1 {2 q_r b} + \frac 1 {2b^2}$

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

However, by hypothesis $\dfrac a b$ 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:

 $\ds \size {\frac a b - \frac {p_r} {q_r} }$ $=$ $\ds \size {\frac {q_r a - p_r b} {q_r b} }$ $\text {(2)}: \quad$ $\ds$ $\ge$ $\ds \frac 1 {q_r b}$

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 by hypothesis $r$ 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 $x$

$\blacksquare$