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Given $0 \lt a,b\in\mathbb{R}, a\ne b$, for all $x, \epsilon$ do there exist integral $n,m$ such that $|\dfrac{a^n}{b^m} - x| < \epsilon$?

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    $\begingroup$ What if, e.g., $a=1, b=2$? Or for that matter, $a=2, b=4$? (I presume you want $m,n$ to be integral...) $\endgroup$ – Steven Stadnicki Dec 28 '19 at 21:12
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    $\begingroup$ My answer isn't right. Please unaccept it. I will work on a better one. $\endgroup$ – John Bentin Dec 29 '19 at 10:24
  • $\begingroup$ Please consider my new answer, which I have checked carefully. $\endgroup$ – John Bentin Jan 1 at 11:11
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Certainly not: if, say, $a=b^l$ with an integer $l>1$, then you want $b^{ln-m}$ to be within $\varepsilon$ from $x$; but $b^{ln-m}$ is either smaller than $1$, or an integer; thus, if $x>1$ and $\varepsilon$ is smaller than the distance from $x$ to the nearest integer, then the inequality in question fails to hold.

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In general, no, as remarked in Steven's comment. But you can do it, for integral $m$ and $n$ and positive $\epsilon$, if $a$ is not a rational power of $b$ and neither equals $1$, all of which we assume here. For convenience, we treat the case $a,b,x>1$; the other cases are similar but with some changes of sign on the way.

By Dirichlet's approximation theorem, there are arbitrarily large positive integers $q$ and corresponding $p$ such that $$\left|\frac{\ln a}{\ln b}-\frac pq\right|<\frac1{q^2},$$so that$$0<\left|q\ln a-p\ln b\right|<\frac1q\ln b,$$the left-hand inequality holding because $a$ is not a rational power of $b$. We consider the case$$0<q\ln a-p\ln b<\frac1q\ln b,$$the other case being similar. Thus, for some $\lambda\in(0\;\pmb,\;1)$, we have$$q\ln a-p\ln b=\frac{\lambda\ln b}{q},$$ and therefore$$\frac{a^q}{b^p}=\exp\frac{\lambda\ln b}q.$$By choosing $q$ sufficiently large, we can make the RHS, and hence the LHS, of the above equation equal a number of the form $1+\varepsilon$, where $\varepsilon>0$ is as small as we like.

Now consider the function $t\mapsto(1+\varepsilon)^t\;$ ($t\in\Bbb R$). This strictly increasing function takes all values in $\Bbb R_{>0}\,$, including $x$. Hence there are consecutive integers $k$ and $k+1$ such that$$(1+\varepsilon)^k\leqslant x\leqslant(1+\varepsilon)^{k+1}$$or$$\left(\frac{a^q}{b^p}\right)^k\leqslant x\leqslant\left(\frac{a^q}{b^p}\right)^k(1+\varepsilon).$$ Thus

$$\frac{a^{qk}}{b^{pk}}\leqslant x\leqslant\frac{a^{qk}}{b^{pk}}+\frac{a^{qk}}{b^{pk}}\varepsilon\leqslant\frac{a^{qk}}{b^{pk}}+x\varepsilon.$$

By taking $\varepsilon< \epsilon/x$, $m=pk$, and $n=qk$, we have the $\epsilon$-approximation sought in the question.

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