Convergents are Best Approximations
Theorem
Let $x$ be an irrational number.
Let $(p_n)_{n\geq0}$ and $(q_n)_{n\geq0}$ 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:
- $\forall n > 1: \left\vert{q_n x - p_n}\right\vert \le \left\vert{b x - a}\right\vert$
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 $1 \le b \le q_n$:
- $\left\vert{x - \dfrac {p_n} {q_n}}\right\vert \le \left\vert{x - \dfrac a b}\right\vert$
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 $b_n$, and the second by $a_n$, then subtracting, we get:
- $a q_n - b p_n = s \left({p_{n + 1} q_n - p_n q_{n + 1} }\right)$
After applying Difference between Adjacent Convergents of Simple Continued Fraction we get:
| \(\ds s\) | \(=\) | \(\ds \left({-1}\right)^{n + 1} \left({a q_n - b p_n}\right)\) | ||||||||||||
| \(\ds r\) | \(=\) | \(\ds \left({-1}\right)^{n + 1} \left({b p_{n + 1} - a q_{n + 1} }\right)\) | 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 $q_{n + 1} \mathrel \backslash b$ 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.
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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 $r \left({q_n x - p_n}\right)$ and $s \left({q_{n + 1} x - p_{n + 1} }\right)$ have the same sign.
This is necessary for the Triangle Inequality to hold.
So:
| \(\ds \left\vert{b x - a}\right\vert\) | \(=\) | \(\ds \left\vert{\left({r q_n + s q_{n + 1} }\right) x - \left({r p_n + s p_{n+1} }\right)}\right\vert\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left\vert{r \left({q_n x - p_n}\right) + s \left({q_{n + 1} x - p_{n + 1} }\right)}\right\vert\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left\vert{r}\right\vert \left\vert{q_n x - p_n}\right\vert + \left\vert{s}\right\vert \left\vert{q_{n + 1} x - p_{n + 1} }\right\vert\) | ||||||||||||
| \(\ds \) | \(>\) | \(\ds \left\vert{r}\right\vert \left\vert{q_n x - p_n}\right\vert\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds \left\vert{q_n x - p_n}\right\vert\) |
as we wanted to prove.
$\blacksquare$