# Condition for Rational to be a Convergent

Jump to navigation
Jump to search

## 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 | ||||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \) | \(<\) | \(\displaystyle \frac 1 {2 q_r b} + \frac 1 {2b^2}\) |

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:

\(\displaystyle \left\vert{\frac a b - \frac {p_r} {q_r} }\right\vert\) | \(=\) | \(\displaystyle \left\vert{\frac {q_r a - p_r b} {q_r b} }\right\vert\) | |||||||||||

\(\text {(2)}: \quad\) | \(\displaystyle \) | \(\ge\) | \(\displaystyle \frac 1 {q_r b}\) |

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.

$\blacksquare$