# Convergents are Best Approximations/Corollary

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## Corollary to Convergents are Best Approximations

Let $x$ be an irrational number.

Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the numerators and denominators of its continued fraction expansion.

Each convergent $\dfrac {p_n} {q_n}$ is a best rational approximation to $x$.

That is, for any rational number $\dfrac a b$ such that $1 \le b \le q_n$:

- $\size {x - \dfrac {p_n} {q_n} } \le \size {x - \dfrac a b}$

The equality holds only if $a = p_n$ and $b = q_n$.

## Proof

Assume otherwise:

- $\exists \dfrac a b$ such that $1 \le b \le q_n$

and:

- $\size {x - \dfrac {p_n} {q_n} } > \size {x - \dfrac a b}$

Then:

\(\ds \size {q_n x - p_n}\) | \(=\) | \(\ds q_n \size {x - \frac {p_n} {q_n} }\) | ||||||||||||

\(\ds \) | \(>\) | \(\ds q_n \size {x - \frac a b}\) | ||||||||||||

\(\ds \) | \(\ge\) | \(\ds b \size {x - \frac a b}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \size {b x - a}\) |

which contradicts the result of Convergents are Best Approximations.

$\blacksquare$