Convergents are Best Approximations

Theorem
Let $\dfrac {p_n} {q_n}$ be the $n$th convergent of the continued fraction expansion of an irrational number $x$.

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 $b = q_n$.

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:

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:

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.

From Relative Sizes of Convergents of Simple Continued Fraction, 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:

as we wanted to prove.