Suppose for contradiction that \(\xi = a/b\) for some integers \(a, b\) with \(b \ne 0\). Then \(b \log 2 = a \log 3\), which implies \(\log (2^b) = \log (3^a)\). Since the exponential function is injective, we must have \(2^b = 3^a\). However, an integer equation of the form \(2^b = 3^a\) holds if and only if \(a = b = 0\), contradicting the assumption that \(b \ne 0\) (or checking via parity for nonzero \(a, b\)). Therefore, \(\xi \) cannot be written as a fraction of integers.

We apply the natural logarithm to the inequality \(1 - \varepsilon {\lt} \frac{3^a}{2^b} {\lt} 1\). The left-hand side gives \(\log (1 - \varepsilon ) {\lt} a \log 3 - b \log 2\), which rearranges to \(b \log 2 - a \log 3 {\lt} -\log (1 - \varepsilon )\). Dividing by \(b \log 3\) (which is positive since \(b {\gt} 0\)), we obtain \(\xi - a/b {\lt} \delta (\varepsilon )/b\). The right-hand side gives \(a \log 3 - b \log 2 {\lt} 0\), which is equivalent to \(a \log 3 {\lt} b \log 2\). Dividing by \(b \log 3\) yields \(a/b {\lt} \xi \), or \(0 {\lt} \xi - a/b\). These steps are reversible.

This is a standard consequence of the theory of continued fractions. The continued fraction convergents of an irrational number \(\xi \) alternate in being strictly less than and strictly greater than \(\xi \). The even convergents \(q_k = p_{2k}/q_{2k}\) provide an infinite sequence of rational approximations strictly smaller than \(\xi \), satisfying the well-known approximation bound \(|\xi - q_k| {\lt} 1 / (q_k)^2\). Since \(\xi \) is irrational, the set of such convergents is infinite.

Since \(\varepsilon \in (0, 1)\), the value \(\delta (\varepsilon )\) is strictly positive. By Lemma 138, there are infinitely many rational approximations \(q = a/b {\lt} \xi \) such that \(|\xi - a/b| {\lt} 1/b^2\). Because there are infinitely many such approximations, we can choose ones where the denominator \(b\) is arbitrarily large. In particular, we restrict to the infinite subset where \(b\) is large enough such that \(b \ge 2\) and \(1/b \le \delta (\varepsilon )\) (or more formally, \(b \ge \lceil 1 / \delta (\varepsilon ) \rceil + 1\)). For such an approximation, we have:

Since \(q {\lt} \xi \), we have \(0 {\lt} \xi - a/b\). Thus, \(0 {\lt} \xi - a/b {\lt} \delta (\varepsilon )/b\). We ensure \(a\) and \(b\) are positive (a trivial consequence since \(\xi {\gt} 0\) and \(b\) is large). By Lemma 137, this implies \(1 - \varepsilon {\lt} \frac{3^a}{2^b} {\lt} 1\). Since the subset constructed in this way is infinite and each fraction maps to a unique pair, there are infinitely many such pairs \((a, b)\).