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Given real $a$ and $b$ with $0<a<1<b$, can every positive real number be arbitrarily well approximated by a number of the form $a^mb^n$ ($m,n\in\Bbb N$), provided that $a^mb^n=1$ only when $m=n=0\,$?

I expect this problem to be well known but am unable to find it by a web search.

Taylor Rendon
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John Bentin
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    This question is equivalent to asking whether the numbers ${m\log a+n\log b}$ are dense in $\Bbb R$, and the "provided that" condition is equivalent to stipulating that $\log a$ and $\log b$ are linearly independent over $\Bbb Q$. At this point the fact that $\log a$ and $\log b$ are logarithms is no longer crucial, and one can search on the group spanned by a set of two linearly independent real numbers. – Greg Martin May 20 '22 at 17:22
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    Since $m,n$ are restricted to be natural numbers, it isn’t true for two positive linearly independent reals. So the equivalent is given two real numbers of opposite sign and linearly independent over $\mathbb Q,$ … @greg – Thomas Andrews May 20 '22 at 17:42
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    Another way to rephrase it is that the condition requires that $\alpha=\log_b(1/a)$ is positive and irrational. Then this is equivalent to saying $$S={m\alpha-n\mid m,n\in\mathbb N}$$ is dense in $\mathbb R,$ since $(1/b)^{m\alpha-n}=a^mb^n.$ You can actually show that $S$ is dense in $\mathbb R$ for any irrational $\alpha.$ – Thomas Andrews May 20 '22 at 17:53
  • Thanks, @ThomasAndrews . If you post your comment as an answer, then I can accept it. – John Bentin May 20 '22 at 17:56
  • More generally, given any $S\subseteq \mathbb R$ which contains $0,$ a negative and positive number, and is closed under addition, is either dense in $\mathbb R$ it is equal to $r\mathbb Z$ for some real $r.$ – Thomas Andrews May 20 '22 at 17:57
  • I suspect this has been asked and answered before, and this question doesn’t match requirements for a “good question,” since no work has been shown, so custom here is to not answer. – Thomas Andrews May 20 '22 at 18:06
  • @ThomasAndrews : Yes, thanks to your comment, I can see that the question is equivalent to this one, which has been well answered. – John Bentin May 20 '22 at 18:35

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