Every Rational Number is Approximable to no Order Greater Than 1

Theorem

Every rational number is approximable to order \(1\) and no greater.

Proof

Since every real number is approximable to order one, we just need to prove that each rational number is approximable to no higher order. Suppose \(\alpha = \frac{a}{b}\) is a rational number, with \(a, b \in \mathbb{Z}\) and \(b > 0\). Then, for each \(p, q \in \mathbb{Z}\) with \(q > 0\) consider that

\[\begin{align*} \left|\alpha - \frac{p}{q}\right| &= \left|\frac{aq - bp}{bq}\right| \\ &= \frac{|aq - bp|}{bq}. \\ \end{align*}\]

If we assume \(\alpha \neq \frac{p}{q}\), then the numerator \(aq - bp\) is non-zero. Since it is a positive integer, it must be at least \(1\) and therefore

\[ \left|\alpha - \frac{p}{q}\right| = \frac{|aq - bp|}{bq} \leq \frac{1}{bq}.\]

Now, assume that \(\alpha\) is approximable to some order \(s\). Thus, using the above inequality as well, there exists a \(c\) such that, for pairs \((p, q)\) with arbitrarily large \(q\) we have

\[ \frac{1}{bq} \leq = \frac{|aq - bp|}{bq} < \frac{c}{q^s}.\]

Therefore

\[ q^{s - 1} = \frac{q^s}{q} < bc.\]

Since neither \(b\) nor \(c\) depend on \(q\), the only way this can be true for arbitrarily large \(q\) is if \(s - 1 \leq 0\), or equivalently that \(s \leq 1\). Thus, \(\alpha\) is approximable to no order greater than \(1\).