Show that between two positive real numbers we can find a number of the form $2^n3^m$,$m,n\in\mathbb{Z}$.
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Answer
Assume that you want to find some number of the form $2^n 3^m$ (with $m,n\in\mathbb{Z}$) in the interval $(a,b)\subset\mathbb{R}^+$. Let we consider the sequence of intervals given by $I_k = (2^k a, 2^k b)$. If one of such intervals contains a power of $3$ (say $3^h$) we are done, since $2^{-k}3^h$ belongs to $(a,b)$.
If we consider the function $\varphi:x\mapsto\log_3(x)$, $\varphi$ maps $I_k$ into
$$ J_k = \left(\log_3(a)+k\log_3(2),\log_3(b)+k\log_3(2)\right) $$
that is an interval with fixed length, $\log_3(b)-\log_3(a)$. We just have to show that for some $k\in\mathbb{Z}$ the interval $J_k$ contains an integer, but that is granted by the irrationality of $\log_3(2)$: it is enough to consider the denominators of convergents of the continued fraction of $\log_3(2)$ that ensure a good enough approximation. To finish the proof, we just need to prove that $\log_3(2)\not\in\mathbb{Q}$. Simple by contradiction: assuming that $\log_3(2)=\frac{p}{q}$ with $\gcd(p,q)=1$, we have $2^q=3^p$, absurd by the unique factorization theorem.
A worked example: since $\log_3(2)\approx\frac{15601}{24727}$, $\;2^{24726}\,3^{-15600}\in\left(1.5,1.51\right).$
To state it shortly, the numbers of the form $a\log 2+b\log 3$ with $a,b\in\mathbb{Z}$ give a dense subset of the real line, and the exponential map is a continuous map, hence it preserves density (in $\mathbb{R}^+$).
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