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\begin{fcprop}[]
  Assuming:
  - $\eps > 0$
  Then:
  there exists $c = c(\eps) >
  0$ such that for $p, q, k, m \in \Zbb_{>0}$, we
  have
  \[
    |p2^k - q3^m| > c \frac{\max(2^k,
    3^m)^{1-\eps}}{\max(p, q)}
  \]
  or $p2^k = q3^m$.
\end{fcprop}

\begin{proof}
  Take $n = 2$, $S = \{2, 3, \infty\}$. Let
  $L_j^{(v)} = X_j$ for all $j, v$, except:
  $L_2^{(\infty)} = X_2 - X_1$. Then the solutions
  of
  \[
    \prod_{v \in S} \prod_{j = 1}^{2} |L_j^{(v)}
    (x_1, x_2)|_v < H(x_1, x_2)^{-\eps / 2}
  \]
  with $x_1, x_2 \in \Zbb$ are contained in the
  lines: $X_1 = 0$, $X_2 = 0$, $X_1 = X_2$ plus
  finitely many points.

  Plug in $X_1 = p2^k$, $X_2 = q3^m$. Then
  \[
    |L_1^{(\infty)}(x_1, x_2)|_\infty = p2^k
    \qquad |L_2^{(\infty)} (x_1, x_2)|_\infty \le
    \frac{\max(2^k, 3^m)^{1 - \eps}}{\max(p, q)}
  \]
  provided $p, k, q, m$ does not satisfy the claim
  with $c = 1$.
  Also,
  \[
    |L_1^{(2)}(x_1, x_2)|_2 \le 2^{-k}
    \qquad
    |L_2^{(2)}(x_1, x_2)|_2 \le 1
  \]
  \[
    |L_1^{(3)}(x_1, x_2)|_3 \le 1
    \qquad
    |L_2^{(3)}(x_1, x_2)|_3 \le 3^{-m}
  \]
  so
  \[
    \eqref{subthm_concleq} \le \frac{p}{3^m}
    \cdot \frac{\max(2^k, 3^m)^{1 - \eps}}{\max(p,
    q)} \le \frac{\max(2^k, 3^m)^{1 - \eps}}{3^m}
  .\]
  Assume $3^m \ge 2^k$ by symmetry. Then
  \[
    \eqref{subthm_concleq}
    \le \max(2^k, 3^m)^{-/eps}
  .\]
  We can assumme that $p, q \le 3^m$, for
  otherwise the claim is trivial. Then $H(p2^k,
  q3^m) \le 3^{2m}$. Then
  \[
    \eqref{subthm_concleq}
    < H(p2^k, q3^m)^{-\eps/2}
  .\]
  Then either $p2^k = q3^m$ or $p, q, k, m$ is one
  of finitely many exceptions.

  Make $c$ small enough to rule out the exceptions.
\end{proof}

For $a, b \in \Zbb_{>0}$, let $N(a, b)$ denote the
number of non-zero digits in the base $b$
expansion of $a$.

\begin{theorem}[Senge, Strauss]
  We have $N(a, 2) + N(a, 3) \to \infty$ as $a \to
  \infty$.
\end{theorem}

Despite the fact that this statement looks quite
modest, the proof is not so simple.

\begin{proof}
  Take $a \in \Zbb$: we assume that $N(a, 2) +
  N(a, 3) < N$ for some fixed $N$. Consider its
  base $2$ expansion.

  First we will explore the consequences of having
  a large string of $0$s in the base $2$
  expansion.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/bf35ef5edb0c4669.png}
  \end{center}
  Then $a = p \cdot 2^{k_1} + e_1$. We know:
  \[
    |p| < 2^{\log_2(a) - k_1 + 1}, \qquad |e_1| <
    2^{k_2}
  .\]
  Similarly: $a = q \cdot 3^{m_1} + e_2$ with $|q|
  < 3^{\log_3(a) - m_1 + 1}$ and $|e_2| <
  3^{m_2}$.

  We will make sure that $\frac{2^{k_1}}{3^{m_1}},
  \frac{2^{k_2}}{k^{m_2}} \in \left[ \frac{1}{3},
  3\right]$.
  \[
    |p2^{k_1} - q3^{m_1}| = |e_1 - e_2| < 3 \cdot
    2^{k_2}
  .\]
  Want to use the proposition. So we need:
  \[
    |p2^k - q3^m| \cdot \max(p, q) < c \max(2^k,
    3^m)^{1-\eps}
  .\]
  So we want
  \[
    C \cdot 2^{k_2} \cdot 2^{\log_2(a) - k_1} < c
    \cdot 2^{k_1(1 - \eps)}
  .\]
  We want
  \[
    \log_2(a) - k_1 < k_1 - k_j - \eps \log_2(a)
  .\]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/ae15bc09bef84775.png}
  \end{center}
  Since at most $N$ blocks have a non-zero number,
  one of the blocks only has zeroes, which can be
  used with the above to show that $a$ cannot be
  too large.
\end{proof}

The constants in all results so far (except
\nameref{liouville}) are \emph{ineffective}!

Are there any improvements of
\[
  \left| 2^{1 / 3} - \frac{p}{q} \right| <
  \frac{100}{q^3}
\]
(suppose $100$ is the best you can get with
\nameref{liouville}) for $q <
10^{10^{10^{10^{10^{10^{10^{10}}}}}}}$. No!

To demonstrate what it means that the above
results are ineffective:

Suppose that we want to find all the solutions of
$x^3 - 2y^3 = 11$. Thue says that we have finitely
many. But because it is ineffective, we have no
idea how to bound the largest of these is, so
would struggle to find all solutions, even with an
arbitrarily powerful computer (or an army of
postdocs).
