Convergents are Best Approximations

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Theorem

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.

Let $\dfrac {p_n} {q_n}$ be the $n$th convergent.

Let $\dfrac a b$ be any rational number such that $0 < b < q_{n + 1}$.


Then:

$\forall n > 1: \size {q_n x - p_n} \le \size {b x - a}$

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


Corollary

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

Let $\dfrac a b$ be a rational number in canonical form such that $b < q_{n + 1}$.

Suppose it is not true that $a = p_n$ and $b = q_n$, in which case the equality certainly holds.

Consider the system of equations:

\(\ds a\) \(=\) \(\ds r p_n + s p_{n + 1}\)
\(\ds b\) \(=\) \(\ds r q_n + s q_{n + 1}\)

Multiplying the first by $q_n$, and the second by $p_n$, then subtracting, we get:

$a q_n - b p_n = s \paren {p_{n + 1} q_n - p_n q_{n + 1} }$


After applying Difference between Adjacent Convergents of Simple Continued Fraction we get:

\(\ds s\) \(=\) \(\ds \paren {-1}^{n + 1} \paren {a q_n - b p_n}\)
\(\ds r\) \(=\) \(\ds \paren {-1}^{n + 1} \paren {b p_{n + 1} - a q_{n + 1} }\) by a similar process

So $r$ and $s$ are integers.

Neither of them is $0$ because:

if $r = 0$ then $a q_{n + 1} = b p_{n + 1}$, and Euclid's Lemma means $q_{n + 1} \divides b$ as $p_{n + 1} \perp q_{n + 1}$, which contradicts $0 < b < q_{n + 1}$
if $s = 0$ we have $\dfrac a b = \dfrac {p_n} {q_n}$ and this we have already excluded as a special case.


From Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent, the convergents are alternately greater than and less than $x$.

Hence since $0 < b = r q_n + s q_{n + 1} < q_{n + 1}$, the integers $r$ and $s$ must have opposite sign.

It follows that $r \paren {q_n x - p_n}$ and $s \paren {q_{n + 1} x - p_{n + 1} }$ have the same sign.

This is necessary for the Triangle Inequality to hold.

So:

\(\ds \size {b x - a}\) \(=\) \(\ds \size {\paren {r q_n + s q_{n + 1} } x - \paren {r p_n + s p_{n+1} } }\)
\(\ds \) \(=\) \(\ds \size {r \paren {q_n x - p_n} + s \paren {q_{n + 1} x - p_{n + 1} } }\)
\(\ds \) \(=\) \(\ds \size r \size {q_n x - p_n} + \size s \size {q_{n + 1} x - p_{n + 1} }\)
\(\ds \) \(>\) \(\ds \size r \size {q_n x - p_n}\)
\(\ds \) \(\ge\) \(\ds \size {q_n x - p_n}\)

as we wanted to prove.

$\blacksquare$