Convergents are Best Approximations

Theorem
Let $$\frac {p_n} {q_n}$$ be the $$n$$th convergent of the irrational number $$x$$.

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

Then:
 * $$\forall n > 1: \left|{q_n x - p_n}\right| \le \left|{b x - a}\right|$$.

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

Corollary
Let $$\frac {p_1} {q_1}, \frac {p_2} {q_2}, \ldots$$ be the convergents of the irrational number $$x$$.

Then for any rational number $$\frac a b$$ such that $$1 \le b \le q_n$$:
 * $$\left|{x - \frac {p_n} {q_n}}\right| \le \left|{x - \frac a b}\right|$$.

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

Proof
Let $$\frac a b$$ 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 Properties of Convergents of Continued Fractions 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 $$q_{n+1} \backslash b$$ as $$p_{n+1} \perp q_{n+1}$$, which contradicts $$0 < b < q_{n+1}$$;
 * if $$s = 0$$ we have $$\frac a b = \frac {p_n} {q_n}$$ and this we have already excluded as a special case.

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

(This is because the convergents are alternately greater than and less than $$x$$ from Relative Sizes of Convergents of Simple Continued Fraction.

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 so we can use the Triangle Inequality.)

So:

$$ $$ $$ $$ $$

as we wanted to prove.

Proof of Corollary
Assume otherwise, i.e. that $$\exists \frac a b$$ such that $$1 \le b \le q_n$$ and $$\left|{x - \frac {p_n} {q_n}}\right| > \left|{x - \frac a b}\right|$$.

Then:

$$ $$ $$ $$

which contradicts the result of the theorem.