Convergents are Best Approximations
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.
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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 $r \paren {q_n x - p_n}$ 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.
$\blacksquare$