Dirichlet's approximation theorem: Difference between revisions

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Suppose ''α'', ''p'', ''q'' are such that <math>\left|\alpha - \frac{p}{q}\right| < \frac{1}{2q^2}</math>, and assume that ''α'' > ''p''/''q''. Then we may write <math>\alpha - \frac{p}{q} = \frac{\theta}{q^2}</math>, where 0 < ''θ'' < 1/2. We write ''p''/''q'' as a finite continued fraction [''a''₀; ''a''₁, ..., ''a_n''], where due to the fact that each rational number has two distinct representations as finite continued fractions differing in length by one (namely, one where ''a_n'' = 1 and one where ''a_n'' ≠ 1), we may choose ''n'' to be even. (In the case where ''α'' < ''p''/''q'', we would choose ''n'' to be odd.)

Let ''p''₀/''q''₀, ..., ''p_n''/''q_n'' = ''p''/''q'' be the convergents of this continued fraction expansion. Set <math>\omega := \frac{1}{\theta} - \frac{q_{n-1}}{q_n}</math>, so that <math>\theta = \frac{q_n}{q_{n-1} + \omega q_n}</math> and thus,<math display="block">\alpha = \frac{p}{q} + \frac{\theta}{q^2} = \frac{p_n}{q_n} + \frac{1}{q_n (q_{n-1} + \omega q_n)} = \frac{(p_n q_{n-1} + 1) + \omega p_n q_n}{q_n (q_{n-1} + \omega q_n)} = \frac{p_{n-1} q_n + \omega p_n q_n}{q_n (q_{n-1} + \omega q_n)} = \frac{p_{n-1} + \omega p_n}{q_{n-1} + \omega q_n}, </math> where we have used the fact that ''p_n''_-1−1 ''q_n'' - ''p_n'' ''q_n''_-1−1 = (-1)''ⁿ'' and that ''n'' is even.

Now, this equation implies that ''α'' = [''a''₀; ''a''₁, ..., ''a_n'', ''ω'']. Since the fact that 0 < ''θ'' < 1/2 implies that ''ω'' > 1, we conclude that the continued fraction expansion of ''α'' must be [''a''₀; ''a''₁, ..., ''a_n'', ''b''₀, ''b''₁, ...], where [''b''₀; ''b''₁, ...] is the continued fraction expansion of ''ω'', and therefore that ''p_n''/''q_n'' = ''p''/''q'' is a convergent of the continued fraction of ''α''.